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■WilUara Collins, Sons, & Co.'h Educntiomil Worka,
■ COLLINS' ELEMENTARY AMD ADVANCED SCIENCE SERIES,
Adabiid If til riqiiirmiili of rAc &,ulh Kinsitiir'^.n Sylhbas, fir Sl-jdinti m
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(SoUias' Jlbbmtxeb Science Seticjs.
PURE MATHEMATICS,
AEITHMETIC, ALGEBEA, GEOMBTBT,
PLANE TBIGONOMETET.
EDWARD ATKINS, B.Sc. (Lond.),
LONDON' AND GLASGOW:
"WILLIAM COLLINS. SONS, Jt COMPANY.
1874
Id^riekUtatmii
/i'7, ^ //
I
TO THE REVEREND
WILLIAM FEY, M.A.,
HONORART CANON OF PETERBOROUGH,
• AND SECRETARY OP THE
LEICESTER ARCHIDIACONAL BOARD OF EDUCATION,
^kxa Moxk i0 ©ffertb.
AS A TRIBUTE OF RESPECT FROM ONE OF HIS
OLD PUPILS.
-.«
PREFACE.
The special object of the present work is to meet the
requirements of the Science and Art Department's Examina-
tions in the first three stages of Pure Mathematics as set down
in the Syllabus of the Science Directory. This will account
for the arrangement of the subject matter. I hope, how-
ever, that it will be found not unsuitable as a general class-
book in Elementary Mathematics.
In the Arithmetical Section my object has been to deduce
the rules from first principles, avoiding as much as possible
algebraical considerations.
The Geometry consists simply of the first three books
of Euclid's Elements with exercises, the only point calling
for remark being the marginal notes. I have found it
useful in class teaching to put down upon a blackboard
the chief steps of the proposition — just those points, in fact,
which it is necessary to retain in the memory ; and to
encourage the pupil to. depend upon himself for supplying
the connecting links. This skeleton, as it were, of the
proposition is placed in the margin, with the hope that
it will be specially appreciated by many students of the
industrial classes to whom the language of Euclid is
ordinarily an insuperable barrier. Particular reference is
here made to such of those classes as have grown up to
almost manhood without any mathematical training.
In the Algebraical Section of Stage I., the more difficult
examples are set particularly for the exercise of thoao
6 PREFACE.
students who take up the work of Stage 11. without ha\'ing
previously used this book for Stage I.
The amount of Plane Trigonometry included in this
volume is small, extending only to the solution of triangles
and the simpler cases of heights and distances. There
is, however, sufficient to fully cover the requirements of
Stage II. of the Government Syllabus. The treatment of the
higher parts of Algebra and Plane Trigonometry, as well
as that of Spherical Trigonometry, is reserved for a second
volume.
E. A.
Leicester, November^ 187$,
CONTENTS.
STAGE L
SECTION I.— ARITHMETIC.
Chap. I. — The Fundamental Pbinciples and Rules Ap-
plied TO Whole Numbebs and Decimal Frac-
tions, 9
,, n. — The Treatment of Frachons considered as
Ratios, 16
„ III. — Application to Concrete Quantities, , . 42
„ IV. — The Metric System, 49
„ V. — Proportion, 61
„ VI. — ^Application to Ordinary Questions of Com-
merce AND Trade, 63
Miscellaneous Examples, 85
SECTION IL— GEOMETRY.
Euclid's Elements, Book L, 89
SECTION III.— ALGEBRA.
Chap. I. — Elementary Principles, . . • • • 141
„ II. — Addition, Subtraction, Multiplication, and
Division, 155
„ III. — ^Involution and Evolution, .... 178
„ IV. — Greatest Common Measure, and Least Common
Multiple, 195
„ v.— Fractions, . , . • • . • 204
„ VI. — SiBfPLE Equations and Problems Producing
them, •••..... 209
8 CONTENTS.
STAGE II
SECTION I.— GEOMETRY.
Euclid's Elements, Book II., .,,.,. 234
Euclid's Elements, Book III.,. . . . • . 254
SECTION II.— ALGEBRA.
Chap. I. — Quadratic Equations, 296
„ n. — Problems Producing Quadratic Equations, . 312
„ III. — Indices, . 316
„ IV.— Surds 319
V. — Ratio and Proportion, 327
99
SECTION III.— PLANE TRIGONOMETRY.
Chap. I. — Modes of Measuring Angles by Degrees and
Grades, . 333
»»
11. — ^The Goniometric Functions, .... 336
„ III. — Contrariety of Sig ns — Changes of Magnitude
AND Sign of the Trigonometrical Ratios
THROUGH the Four Quadrants, . . . 342
„ IV. — Trigonometrical Ratios— continued. — Arith-
metical Values of the Trigonometrical
Ratios of 30°, 45^ 60% &c., . . . 346
„ V. — Logarithms, . . . , . . . 354
„ VI. — The Use of Tables, 358
„ VII. — Properties of Triangles, .... 366
,,VIII. — Solution of Right- Angled Triangles, . 374
„ IX. — Solution of Oblique- Angled Triangles, . 377
• „ X. — Heights and Distances, .... 381
Answers, 387
MATHEMATICS.
FIRST STAGE.
SECTION L
ARITHMETIC.
CHAPTER I.
THE FUNDAMENTAL PRINCIPLES AND RULES APPLIED
TO WHOLE NUMBERS AND DECIMAL FRACTIONS.
Notation and Numeration.
1. We leam from elementary books on Arithmetic, that
figures have a local as well as an intrinsic value, and that
the local value of a figure increases tenfold, or diminishes
tenfold, according as its position is changed from right to
left, or from left to right. Thus, commencing with the right
hand figure of an ordinary number, the respective figures of
the number stand for units, tens, hundreds, thousands, &c. ;
or, beginning with the left hand figure, which, we will
suppose, stands for thousands, the respective figures repre-
sent thousands, hundreds, tens, units. Let us carry this
principle a little further. Take the figures 68754, and
suppose that 7 represents 7 units; the question then arises
as to the number represented by 68754. Now, as 7 is the
units* figure, we have evidently, by the above principle, 6
hundreds, 8 tens, 7 units; and further, remembering that the
10 ARITHMETIC.
local value of a figure decreases tenfold for every remove to
the right, the 5, on our supposition, must represent 5 tenths,
and the 4 must represent 4 hundredths. Let us, as is
usual in numbers thus represented, mark the units' figure by
placing a dot to the right of it. Thus, 357*2605 will then
represent 3 hundreds, 5 tens, 7 units, 2 tenths, 6 hundredths,
5 ten-thousandths; and to take one other example, '3065,
where the units* figure, though not expressed, is actually 0,
will represent 3 tenths, 6 thousandths, 6 ten-thousandths.
The dot is called the decimal point, and the digits to the
right are called decimals, because they represent portions of
the iinit obtained by cutting it up into a number of equal
parts, wliich is always some power of 10. It may be
remarked, that 10 is called the first power of 10; 100,
or 10 X 10, the second power, sometimes written 10^;
1000, or 10 X 10 X 10, the third power, written 10^, and
so on.
To make the subject clear, let us see what the decimals,
•237, '2370, -0237 respectively represent. Now, the digits
2, 3, 7-, in the first two decimals, are in exactly the same
position with regard to the decimal point, and the respective
digits in each have the same absolute value; moreover, the
cipher affixed to the right of the decimal '2370, has no
intrinsic value, and hence the two decimals, -237, and -2370,
have the same absolute value. And since the reasoning is
the same, no matter how many ciphers are affixed to the
right, we get the following important principle : —
The value of a decimal is not altered by affixing ciphers
to the right.
We will now compare the first and third of our examples,
namely, the decimals -237 and -0237. The cipher which is
hero jyrejixed to the left, has again no intrinsic value ; but it
has removed the digits 2, 3, 7, one stage to the right, and
has, therefore, diminished their local value tenfold. The
effect of prefixing the cipher, is therefore to diminish the
absolute value of the decimal tenfold, and as every additional
cipher so prefixed has a similar effect, we get another funda-
mental principle, as follows : —
The value of a decimal is diminished tenfold for every
cipher prefixed.
ADDITION AND SUBTEACTION. 11
After the above it is easy to see that we have only to
remove the decimal point one place to the right or left, in
order to increase or diminish respectively the value of a
decimal tenfold. And if we allow the term decimal to in-
clude numbera which are greater than unity, as 35*721, we
may extend the principle thus : —
A decimal may be divided by 10, by removing the dot one
place to the left; by 100 or 10^ by removing the dot two
places to the left; by 1000 or 10' by removing the dot three
places to the left, and so on ; and further, a decimal may be
multipHed by 10, 100, 1000, &c., that is, by 10, 10», 10^, Ac,
by removing the dot 1, 2, 3, «fec., places respectively to the
right. Thus, 6872-3476 divided by 10, 100, 1000 respec-
tively, becomes 687*23476, 68*723476, 6*8723476 ; and mul-
tiplied by the same becomes 68723-476, 687234*76, 6872347*6
respectively.
Addition and Subtraction.
2. If the student has understood the preceding article, he
will at once perceive that, provided we keep the units* figure
under the units' figiire in every case, there is no difference
between the addition and subtraction of decimals, and the
addition and subtraction of ordinary integers. All he has to
take care of is that the decimal points are kept under each
other.
Ex. 1.— Add together 32502, *647, 5*6073, *00214, 290,
and 4*7001. Proceeding as in ordinai*y addition :
325*02
•647
5*6073
•00214
290-
4*700 1
625*97654
Ex. 2.— Take 6*291 from 18-3064. Proceeding as in
ordinary subtraction :
18*3064
6*291
120154
13 ARITHMETIC.
Ex. 3. — ^Find the difference between 15*02 and -6732.
15-02
•673 2
14-3468
Note. — In an example of this kmd, where the number of decimal
fibres in the lower line exceeds the number in the upper, it is ad-
visable to mentally supply ciphers to make up the deficiency in the
upper line. This may be done, as we have seen, without altering
the value of the upper line.
Multiplication.
3. Suppose we have to multiply 2*935 by 6*34, and let us
suppose the dot in each case removed to the extreme right.
Then (Art. 1), we have multiplied the number 2935 by 1000.
and the number 6*34 by 100, and we have obtained the
numbers 2935- and 634- respectively. As these numbers are
integers, we may omit the dot, and write them 2935 and
634. Now 2935 x 634 = 1860790, but as we increased our
original numbers one thousand and one hundredfold respec-
tively, it is evident that our product is increased 1000 x 100,
or one hundred thousandfold. Dividing, therefore, the
above result, 1860790 by 100000, or what is the same thing
(Art. 1), writing it 1860790* and removing the dot 5 places
to the left, we get for our product of the numbers 2*935
and 6*34 the result, 18*60790. We may remark that the
number of decimal figures in the product, namely, 5, is the
sum of the numbers of decimal figures in the two given numbers.
We have, therefore, the following rule for multiplication : —
Multiply the given numbers exactly as integers, regardless
of the decimal points, and after the operation is finished,
point off as many decimal figures in the product as there
are together in the multiplier and multiplicand.
Ex. 1.— Multiply 6*35 by -1703.
•1703
6*35
8515
5109
10218
1081405
DIVISION. 13
Now, the number of decimal figures in the multiplier and
multiplicand together, is (4 + 2), or 6, and therefore we
mark oflf 6 decimal figures in our product. This gives us
1 -081405.
Ex. 2.— Multiply -0063, by -017.
•0063
'017
441
1071
And pointing off (4 + 3), or 7 decimal figures, we obtain for
our product -0001071.
Division.
4. Suppose we have to divide -76875 by 6*25. "We will
proceed as in the case of multiplication, by imagining the
decimal points in each number removed to the extreme right.
The numbers will then be 76875*, and 625*, or, omitting the
dot, as they are now integers, they will be 76875, and 625.
Pi*oceed now as in ordmary division (which operation it is
unnecessary to explain), and we get for our quotient 123.
Now we must remember that we have increased our divi-
dend 100,000 or lO'^-fold, and that, consequently, our quotient
will require to be diminished lO'-fold. This is done (Art. 1)
by removing the dot of the number 123- five places to the
left. But, before doing that, we know that the divisor has
been increased 100 or 10*-fold, and on this account our
quotient must be increased 10*-fold. This is done (Art. 1)
by removing the dot two places to the right. Hence, to get
the true quotient of -76875 by 6*25, we must remove the dot
from the extreme right of the number 123*, five minus two,
or three places to the left. Now three is the excess of the
nimiber of decimal figures in the dividend over the number
in the divisor. We hence arrive at the following jule : —
Proceed as in ordinary division, and when all the figures
of the dividend have been brought down, and the remainder,
if any, obtained^ cut off as many decimal figures in the
14 ARITHMETIC.
quotient, as the number of decimal figures in the dividend
exceeds the number in the divisor.
Note. — ^When th6 number of decimal figures in the dividend is less
than the number in the divisor, affix a sufficient number of ciphers
to make the number of decimals in the dividend equal to the number
in the divisor. After finishing the operation of ordinary division,
there will be no decimal figures to cut oflF in the quotient. If there
be a remainder, and the division carried on further, by affixing
ciphers to the successive remainders, all the quotient figures thus
obtained will be decimals.
Ex. 1.— Divide 117-85088 by 6-272.
6-272)117-85088(1879
6272
55130
50176
49548
43904
56448
56448
We see that there are five decimal figures in the dividend,
and three in the divisor, and so we cut off* (5 - 3) or 2 in
the quotient. The answer is therefore 18-79.
Ex. 2.— Divide 527-2 by -0008.
Here it will be necessary to affix three ciphers to the
dividend, and the operation will stand thus —
'0008) 527-2000
659000
As there is no remainder, and the number of decimal
figures in the dividend is equal to that in the divisor, we
have none to cut off. The answer is therefore 659000.
Ex. 3.— Divide 463*7 by 2-769 to four places of decimals.
Here we must affix two ciphers to the dividend, and the
opei-ation, as far as the ordinary remainder of long division,
stands thus : —
DIVISION. ] 5
2-769)463-700(167
2 769
18680
16614
•20660
19383
'1277"
The quotient up to this point is integral ; but, as "we have
a remainder, we must continue the operation of division,
first placing a dot at the right of the figures in the quotient,
and affixing a cipher to the present and each successive re-
mainder, until we have the requisite number of decimals in
the quotient. By thus proceeding, it is easy to see we
arrive at an answer — 167*4611.
Ex. I.
1. Increase the numbers 4-623, 29, -02367, '07 respectively
10, 100, 1,000, 10,000-fold.
2. Divide by inspection the numbers 0-06, 1111, 4*0020,
45 respectively by 100, 10,000, 1,000, 10.
3. Express in words -3467, 34*67, -0003467, 3*467 ; and
compare the values of the last three with the first.
4. Add together —
(1.) 6*732, 14*9, -0064, 14*27006.
(2.) -00291, -291, 29, 29100*9.
(3.) -821, 29*60, 29*6, *0029.
^ 5. By how much does 5 exceed 4*2763, and 16*021 exceed
1^2*700091
6. Find the value of —
(1.) 74*25 + *0067 - 3*0298 + 1*032 - 2*73.
(2.) 3*276 - 8*2409 + 100326 - -00091.
(3.) 2*5 - -00029 - 7*364 + 5*2791.
7. What number added to four thousandths will give three
feundredths, and what number subtracted from 8,000 units
^rill give 291 units 29 hundredths?
16 ABITHMETIC.
8. Find value of —
(1.) -3 X -3. (2.) 9001 X 27-06. (3.) 0403 x -009.
(4.) 17 X 017 X 100. (5.) -3 X -005 x 64.
(6.) {'4:y X (032)1
9. Find the quotient of —
(1.) 79-4 by -397. (2.) 5-928 by 4742-4. (3.) 28
by -007. (4.) -6426 by 2-8. (5.) (•24)« by 9-6.
(6.) 1-806 by (1-9)1
10. Given the quotient -00073, the dividend 124-1, find
the divisor when there is no remainder.
11. What is the value of —
(1.) )2 - -815| -r { -201 + -039 - -0021?
(2.) l(-693)' - (•307)»f -r {-693 - -307}?
12. If I add '061 to a certain number, and then divide
the result by 290, I get -0009 for a quotient ; what is the
number ?
CHAPTER 11.
THE TREATMENT OF FRACTIONS CONSIDERED AS RATIOS.
5. A Fraction is a part or parts of a whole. It is gener-
ally expressed by two numbers, the one placed above the
other and separated by a line. The lower number expresses
the number of equal parts into which the whole quantity has
been divided, and the upper number, how many of those
parts are taken. Thus f is a fraction, and tells us that
unity has been divided into 5 equal parts, and that we have
taken 3 of those parts. The fraction | is read three fifths ;
each of the equal parts into which imity has been divided
being called a fifth.
The denominator of a fraction is the lower number, and
therefore shows the number of equal parts into which we
have divided the imit.
The numerator is the upper number, and tells us how
many of these equal parts are taken.
When the numerator is less than the denominator, the
quantity expressed is actually less than a whole. The
quantity is therefore a real or proper fraction. Again, when
TREATMENT OF FRACTIONS CONSIDERED AS RATIOS. 17
the numerator and denominator are both integral numbers,
the fraction is termed a simple fraction.
Thus f , J, \^, are both proper and simple fractions.
It is however usual to include in the term fraction every
expression which contains one or more simple fractions, with
or without integral numbers.
Thus }, i, VS ^h 3-y P' ^i of I of ^i, axe all included
in the term fraction.
They are, moreover, called vnlgax fractions to distinguish
them from decimals, which, as will be shown further on,
may be looked upon as fractions, according to the above
definition, whose denominators are powers of 10, and not
expressed but understood.
It is convenient to classify fractions as follows : —
(1.) A proper fraction is one whose numerator is less than
its denominator, as f, |, -Jf , -^,
(2.) An improper fraction is one whose numerator is not
51
less than its denominator, as |, f , -Jl, ^.
(3.) A simple fraction is one whose numerator and denom-
inator are both integral numbers, as |, f , x^«
(4.) A mixed number is a fraction expressed by an integer
and a simple fraction, as 2^, 4|, 3f .
(5.) A complex fraction has its numerator, or denominator,
or both, in a fractional form, as --, -K -I .
' 5^. 6^ 11
(6.) A compound fraction is a fraction of a quantity whicU
3
is itself fractional, as ^ of 2 J, 2^ of 6 1^ of -—.
6. In the preceding article we have spoken of fractions in
the ordinary way. We will now approach them from a
different point of view.
By the term ratio we understand the result of the com-
parison of two quantities with regard to magnitude. There
are two kinds of ratios — ratio by difference or subtraction^
5 B
18 ARITHMETIC.
and ratio by quotient or division. Tlius we may consider
how much one quantity exceeds another, or we may consider
how many times one quantity contains another. The former
kind is called the arithmetical ratio, and the latter the
geometrical ratio. We shall speak only of the latter.
Definition. — ^The ratio between two quantities is that
multiple, part, or parts which the former is of the latter.
It is evident that a ratio can exist only between quantities
of the same kind ; thus, we may compare 12 horses and 6
horses, but not 9 men and 4 miles. And if the quantities
are reduced to the same denomination, we may treat the
quantities as abstract, just as we find the quotient of one
concrete quantity by another, by reducing them both to the
same denomination, and dividing as if tiiey were abstract
quantities.
Now, according to what has been stated above, the ratio of
12 to 7, or, as it is usually written, 12 : 7, is obtained by
dividing 12 by 7 ; and this is the same thing as dividing imity
or 1 into 7 equal parts, and computing how much 1 2 of such
parts amount to. It hence follows that the fraction \f is
properly expressed by the ratio 12:7.
The first term of a ratio is called the antecedent, and the
second is called the consequent ; and hence we may consider
a fraction as a ratio, the nimierator being the antecedent of
the ratio, and the denominator the consequent.
"When the antecedent is equal to the consequent, the ratio
is said to be a ratio of equaHty ; and it is said to be a ratio
of less or greai^ inequality according as the antecedent is less
or greater than the consequent.
Thus, 6 : 6 is a ratio of equality.
3 : 4 is a ratio of less inequality.
11 : 9 is a ratio of greater inequality.
The student will therefore have no difficulty in assenting
to the following definitions : —
1.) A proper fraction is a ratio of less inequality.
^2.) An improper fraction is a ratio of equaUty or of
greater inequality.
(3.) A simple fraction is a ratio whose terms are integers.
Thus, I = 3 ; 5 is a simple fraction.
%
TREATMENT OP FRACTIONS CONSIDERED AS RATIOS. 19
(4.) A mixed number is a ratio of greater inequality, whose
antecedent has been actually divided by its consequent, and
the re^ylt expressed as an integer and simple fraction.
Thus, 2f = V = 17 : 7.
(5.) A complex fraction is a ratio, whose antecedent, or
consequent, or both, are not integers.
Thus, ^ 1, y^ = respectively to Sj : If, 2 : 7J, 9| : 5
If 'i 5
are complex finctions.
(6.) A compound fraction is nn expression containing two
or more ratios to be compounded together.
Thus i of i contains the ratios 3 : 4 and 7:9 to be
compounded; J of ^Y of -^ contains the ratios 2^ : 7,
7 6^ y
3 : 6 J, 7^ : 9 to be compounded.
7. A fraction wJiose mtmercUor and denominator are mul-
tiplied or divided by the aamfie quantity is not altered in value.
Suppose, for example, we multiply the numerator and de-
nominator of the fraction f each by 4, we get f = ^ J. Now
the ratio of 3 I 7 is, from the definition of a ratio, four times
as small asthe ratio (3 x 4) : 7 or 12 I 7; and the ratio 12 : 7
is, for the same reason, four times as great as the ratio
12 : (7 X 4) or 12 : 28. It therefore follows that the ratio
3 : 7 is exactly equal to the ratio 12 I 28, and consequently
Again, suppose we divided each term of the fraction ^^ by 3,
we get 1^ = i. Now the ratio of 15 1 27 is, from the definition
of a ratio, three times as great as the ratio (15 -f 3) .' 27 or
5 I 27 ; and, again, the ratio 5 I 27 is three times as small as
the ratio 5 I (27 ^ 3) or 5 : 9. It therefore follows that the
ratio 15 I 27 = the ratio 5 : 9, and consequently ^^ = ^.
Cor. — ^An integer may be expressed as a fraction with any
given denominator.
For we may consider an integer as a ratio whose conse^
quent is 1, and we may multiply each term of this ratio by
any given number witi^out altering its value.
20 ABITHMETIC.
Thus— 6 = -f = 5-^^ = ^.
Or, 6 = 4 = ^-^ = A4.
8. To multiply a fraction by a whole number, we may
either multiply the numerator by the number, or divide the
denominator by it.
Ex. 5. X 3 = ?-^^ = ^ ^ ; or we may proceed thus —
' f X 3 = -?- = £.
" 9 -i- 3 3
As to the first method —
The ratio 8 \ 9 will be evidently increased three times if
we multiply its antecedent by 3 ; this follows from the defi-
nition of a ratio. We thus get the ratio (8 x 3) ; 9 or 24 ; 9.
It therefore follows that f x 3 = V*
As to tJie second method —
The ratio 8 ! 9 will be evidently increased three times if
we make the consequent three times as small ; and we thus
get the ratio 8 I (9 -j- 3) or 8 ! 3 ; and hence it follows that
* X 3 = |.
It may be remarked that the two results, *,* and f , are of
exactly the same value (Art. 7), since the latter may be ob-
tained from the former by dividing each of its terms by 3.
In actual practice we sometimes pursue the first method,
and sometimes the second. If the denominator of the given
fraction contains the multiplier as a factor, it is more con-
venient to use the second method, thus : —
iji- x 3 = —1-1- = li.
13 1 2 -J- 3 4
On the other hand, when the denominator does not contain
the multiplier as a factor, we use the first method, thus : —
(1.) 7x5= y^l-» = 3^.
^ ^ \Z 13 13
(2.) »- X 10 = S-il-3-2.
Here we see that the numerator and denominator have a
common factor 5, and therefore, by Art. 7, if we divide them
both by it, we have : —
■S\ X 10 = »JL1 .= tf
TREATMENT OP FRACTIONS CONSIDERED AS RATIOS. 21
9. To divide a fraction by a whole number, we may either
multiply the denominator by the number, or we may divide
the numerator by it.
Ex.— Divide if by 4.
We may proceed thus, (1.) if -r 4 = i|±^ = 1*^.
V-*-/ TT • * — 17X4 6B'
As to tJie first method —
The ratio of 12 : 17 will evidently be diminished 4 times
if we divide its antecedent 12 by 4. "We thus get the
ratio (12 -r- 4) : 17 or 3 : 17 ; and it therefore follows that
if - 4 = tv.
As to the second metJiod —
The ratio of 12 : 17 can also be diminished 4 times by
increasing its consequent or divisor 4 times, so that we thus
get the ratio 12 : (17 x 4) or 12 : 68. It therefore follows
that if ^ 4 = -ii.
It may be remarked, as in Art. 8, that the two results, -j^
and ^1, have exactly the same value, for the latter can be
obtained from the former by multiplying each of its terms by
4 (see Art. 7).
And again, in actual practice, we usually take the first
method when the numerator contains the divisor as a factor,
but not otherwise. Thus —
(1.) if -f 6 = ia±<i = ^\.
(2-) H - 5 = rii^ = H-
(3.) if - 8 = ^^5-.
Here it is convenient to divide the numerator and denomi-
nator by the common factor 4 (Art. 7).
We then have if - 8 = -^^—, = — ^ = A-
10. To reduce a mixed number to an improper fraction.
Looking at our definition of a mixed number (J^, 6), the
following rule is evident :
Multiply the integral part by the denominator of the
fractional part, and add in the numerator ; this gives the
required numerator, and the denominator of the fractional
part is the required denominator.
22 ARITHMETIC.
Ex. — Reduce 5 J, 7| to improper fractions.
(1.) 5j- = A2i|±« = Y-
(2.) 7| = ^^4^ = V-
11. To reduce a complex fraction to its equivalent simple
fraction. Before stating a rule, let us take an example.
Suppose we liave to reduce -^ to an equivalent simple
oj-
fraction.
H _ i2iA±x V
.6
Now, by the last Art., rr = •^— = . ,-
9
Again, the ratio V : V "^1 ^o* he altered in value if we
multiply both its terms by the same quantity. Let us multiply
them by 9 and it becomes V* x 9 : V ^9* Now, by Art.
8, M X 9 = .^«^» and V x 9 = -ai. = V = 47. Theratio
' ® 6 " 9-J-9
then becomes i^^ : 47. We will again multiply the terms
of this ratio by the same quantity, viz. by 5, and we get the
ratio L5JL9 X 5 : 47 X 5. Now, by Art. 8, i^ x 5 =
ieJLg. = i6X9 = 16 X 9. Hence the ratio V • V is ^^^^i-
vaient to the ratio 16 x 9 : 47 x 6, and hence the
fraction i£ = ^^'
4 7 4 7 '^ 8
Now 16 and 9 are called the extreme terms of the complex
fraction ^, and 5 and 47 are called its mean terms.
V
We arrive then at the following rule : —
KuLE. — Bring the numerator and denominator to the form
of simple fractions, then multiply together the extreme terms
for a new numerator, and the mean terms for a new denomi-
nator.
9 9
12. To reduce a compound fraction to its equivalent
simple fraction.
Let it be required to find the simple fraction equivalent
to the compound fraction | of 4.
1*B£ATM£NT of fractions CONSIDERED AS RATIOS. 2^
Now f of ^ is the ratio 3 : 4, where the unit of this ratio
is f . It is therefore, from the definition of ratio, equal to 3
times this unit divided hj 4.
Now 3 times ^ = ^ x 3 = »-^ (Art. 8),
And /. 3times4 -r 4 = *-^U V 4 = ^""-^ = *-^X
^ 7 7X44X9*
Hence we arrive at the result —
I of ^ = ^"Li,
4X7
And in the same way we might show that
X of * of 4 of 11 = ZJ^-iJL^JU.
° ** * 8X9x5x1
Hence the rule : —
Rule. — Multiply together the several numerators for a
new numerator, and the several denominators for a new
denominator.
Ex. 3J of 2i- of 6 = 3JL«JlA of aJLaJLj of f =
V of f of i = ^
6X6
8 xa X 1'
Ex. rr.
1. Reduce the following to improper fractions-
Si, 4J, lOO^V, 35if, IpigV, llH.
2. Keduce the integer 19 to sixths, tenths, thirteenths,
eighteenths, nineteenths, and twentieths.
3. Bring the following fractions to integers, and reduce
them respectively to fourths, sixths, eighths, tenths, twelfths,
and fourteenths —
V, ih V, VA^, Y, V.
4. Multiply the following fractions each by 10, 11, 12 —
h t\> a, 1|, 7i, di.
5. By how much does 8 times the fraction y exceed the
quotient oiy^fhyZl
6. Divide the following fractions each by 6, 7, 18 —
V, ih 1|, h\> 103|, llA.
7. Diminish the following ratios respectively 6, 7, 8-fold —
12:5, 3^3^:4, 9^:2^.
24 . ARITHMETIC.
8. Simplify the expressions —
(1.) J of -J of f of If. (2.) i of 12 of \i of A-
(3.) T^ of 3| of ^ of li. (4.) ii of 2 of V of ^tV
of I of f . (5.) 2i of J of H of 4^^.
9. Reduce to simple fractions —
U n 6i_ofH _2i_ 2i-of3 7i-f3| .
2i'Z^' 3^ '7^of3i|'Hof3j' " 4f ^'
10. Wliat is the difference between 2f of 3i^ of 2j|
and?i.
11. Find the integral^ value of fi^^L:^-?!^.
" S^of f +-rVof4|
4-4- 8i
12. Add the sum of the fractions —5, — *- to their
difference.
13. To reduce a fraction to its lowest terms.
Def. a fraction is in its lowest terms when its numerator
and denominator have no common factor, or are prime to
each other. (Among such common factors, we include either
the numerator or denominator itself, when one of them
happens to be a divisor of the other.)
"When the numerator and denominator have a common
factor, we may divide them both by it (Art. 7) without
altering the value of the fraction. Now, the highest common
factor of two or more numbers is called their greatest common
measure, usually written G.C.M.
Hence we have the following rule : —
KuLE. — To reduce a fraction to its lowest terms, divide
the numerator and denominator by their G.C.M.
Ex. — Reduce f| to its lowest terms. .
We can easily see that 12 is the G.C.M. of 24 and 84 ;
hence, dividing each by 12, we get —
2_4 = 2^ "^ ^ ^ = -I the fraction required.
84 84 — 12 '' ^
14. It is not, however, always easy to tell by inspection
the G.C.M. ; but, before giving a general method for deter-
mining it, it will be useful to make a few remarks as to the
divisibility of numbers in certain cases.
TREATMEKT OB* FRACTIONS CONSIDERED AS RATIOS. 25
A number is divisible as follows : —
By 2, when it is even.
By 3, when the sum of its digits is divisible by 3.
By 4, when the number formed by the last two figures is
divisible by 4.
By 6, when it ends in 5 or 0.
By 6, when it is even, and is also divisible by 3.
By 8, when the number formed by the last three figures is
divisible by 8.
By 9, when the sum of its digits is divisible by 9.
By 10, when it ends in 0.
By 11, when the sum of the digits in the odd places (that
is, the sum of the 1st, 3rd, 5th, &c.) is equal to the sum of
the digits in the even places, or the one exceeds the other by
a multiple of 11.
By 12, when the number formed by its last two figures is
divisible by 4, and the sum of its digits is a multiple of 3.
We may add also : —
(1.) A number is divisible by 37, when it is composed of
digits which are repeated three times, or any multiple of
three times, as 111, 333, 444444, &c.
(2.) A number which has three figures repeated in the
same order is divisible by 7, 11, 13.
Thus 271271, 166165, 23023 are divisible by 7, 11, and
13 ; for the last may be written 023023.
(3.) A number which has four figures repeated in the same
order is divisible by 73 and 137.
Thus 53245324, 2760276 are both divisible by 73 and 137;
for the last may be written 02760276.
Hence a fraction may often then be reduced to its lowest
terms by gi*adually striking out factors determined by in-
spection.
Ex. — Reduce ^/^\ to its lowest terms.
Now 792 and 2244 are each divisible by 4, for the num-
bers 92 and 44, which are formed by the last two figures of
each, are evidently so. Hence, dividing numerator and de-
nominator by 4, we have jlb-?. = _^9 2_-^_4 = j_9n . ggxjh
*' 2 2 44 2244-r4 66l>
is evidently divisible by 3. Hence -^^-^-^ = J " » "^ ^ = 5-«- .
*' *' 3324 »6lT-a Va-l »
26 ARITHMETIC.
and eadi is now evidently divisible by 11. Hence
792 — 66 -^ 1 1 — .6.^
3344 187-rll 17*
Eemabx. — ^It may sometimes happen that, although we are able to
tell by inspection some of the factors of numerator and denominator,
none of them are common to both numerator and denominator. We
cannot then strike them out, but we may use them to determine
what would be left supposing they are struck out, and we may thus
often come upon the G.C.M. of both numerator and denominator.
Ex. — ^Reduce ^\% to its lowest terms.
ISTow 474 is even, and therefore divisible by 2 ; thus
474 ^ 2 = 237. Again, 2133 is divisible by 9, for the sum
of its digits, viz., (2 + 1 + 3 + 3) or 9 is so divisible ; thus,
2133 + 9 = 237.
We have thus learned that, although 2 and 9 are not com-
mon factors, 237 is a common factor, and, in fact, the
G.C.M. Dividing the numerator and denominator of the
given fraction by 237, we get -^-2^ = ^-J^-^^^^f = J.
O tf f O 2183 3133 — 837 g
We proceed now to give the general method of determin-
ing the G.C.M.
16. To find the G.C.M. of two numbers.
BuLE. — ^Divide the greater number by the less, and if there
be no remainder, the less is the G.C.M.; but if there be a
remainder, make a divisor of this remainder, and a dividend
of the first divisor ; if there be a remainder again, make a
divisor of it, and a dividend of the preceding divisor, and so
on until there be no remainder. The last divisor will be the
G.C.M.
Ex.— Find the G.C.M. of 282 and 799.
The operation will stand thus —
282)799(2 ortiius— 282 799 2
564
236)282(1
235
47)236(5
235
The G.C.M. is therefore 47.
282
799
564
235
282
235
47
235
235
•
^n^fiAfMEXI? 0£^ ABACTIONS CONdtDEtUCD AS lULTIOS. 27
The reason of this rule is easy to see. 47 divides 235, and
it therefore divides 235 + 47 or 282. Hence it is a common
divisor of 282 and 235, and it is therefore a common divisor
of 282 and 282 x 2 + 235 or of 282 and 799.
It is, moreover, the highest common divisor; for every
number which divides 799 and 282 must also divide 799 -
282 X 2 or 235 ; and hence every number which is a measure
of 799 and 282 is also a measure of 235 and 282. Similarly,
every number which divides 235 and 282 will also divide
282 - 235 or 47, and hence every measure of 799 and 282 is
a measure of 47. But no higher number than 47 can divide
47, and therefore 47 is the G.C.M of 799 and 282.
16. It is now easy to see how any fraction may be reduced
to its lowest terms.
Ex. — Reduce |Jf to its lowest terms.
148
703
592
111
148
111
37
111
111
4
1
3
Hence 37 is the G.C.M, and dividing
numerator and denominator by it, we get
17. To find the G.C.M. of more than two numbers.
The following rule needs no explanation: —
Rule. — ^Find the G.C.M. of any two of the numbers, then
the G.C.M. of this result and the third number, and so on.
The last result will be the G.C.M. required.
Ex.— Find the G.C.M. of 282, 987, 658, 1128.
The operation will stand thus —
282
141
987
846
141
2 T4~
47
658
564
4
1
2
47
1
1128
94
282
282
141
94
188
188
94
94
24
Hence the G.C.M. is 47.
28 ARITHMfiTlO*
Ex. III.
1. Resolve into prime factors 44, 64, 150, 252, 1269, 462.
2. Find the prime factor of 1386, 1720, 4608, 21175,
15972, 14256.
3. A number is said to be a perfect number when it is
equal to the sum of its aliquot parts. Show that the follow-
ing are each perfect numbei-s — 6, 28, 496, 8128.
4. Show that 284 and 220 are a pair of amicable numbers;
that is, such a pair that each is equal to the sum of the
divisors of the other, unity being here counted as a divisor.
5. Reduce, by inspection, to their lowest terms, the follow-
ing fractions :
(!•) l%y ih ih in, ^%%%, t¥A.
(2.) -i^g^y, tVi^, Hh mh Uh iSlf.
(3.) Hh U%y tVA, iH> iUih TiHfi-
6. Find the G.C.M. of
(1.) 304, 323. (2.) 413, 448. (3.) 377, 533. (4.) 1866,
2832. (5.) 1189, 1517. (6.) 4374, 5103. (7.) 168,
378, 602. (8.) 539, 616, 792. (9.) 780, 1092, 2145.
7. Reduce to their lowest terms the following fractions —
(1.) -i^bVV, t^/tV, irVVV, l*M, HHy iUh
(2.) illf, Hh im, tVA, HUy HH-.
8. Show, without applying the rule for the G.C.M., that
im = ih and that |-|f Jf f = ih
9. It rains in a certain district 634 days out of every
2,219 ; express this fact in the simplest way possible.
10. Out of 1,659 men engaged in a battle, only 1,185
answered the roll-call in the evening ; express by a ratio in
its simplest terms the number missing in relation to the
whole.
11. There are 40 numbers less than 100, and prime to it;
what are they?
12. A, B, C can do a piece of woi-k respectively in 318,
477, 795 hours; express the relative value of A, B, C as
workmen by the simplest integral numbers possible.
THE LEAST COMMON MULTIPLE. 29
The Least Cominon Multiple.
18. It is often necessary to express fractions as equivalent
fractions, having a common denominator; and it is, more-
over, convenient to have this denominator as small as possible.
Now, tliere are always an infinite number of numbers which
will contain each of the given denominators as a factor, and
our problem is therefore to obtain the least of such numbei^s.
Def. — The least common multiple (L.C.M.) of two or
more numbers is the least number which contains each of
the given numbers exactly.
KuLE. — Arrange the numbers in a line, putting one of
them as a divisor. Strike out the greatest factor common to
this divisor, and each of the numbers separately, and place
the several quotients in the line below ; at the same time
bring down every number prime to the divisor. Repeat the
operation upon the second line, and so on until we have a
line of numbers prime to each other. Multiply the several
divisors and the numbers in the lowest line together, and
their continued product will be the least common multiple.
Ex. 1.— Find the L.C.M. of 12, 16, 36, 45, 60.
36
1
12, 16, 36, 45, 60
5
1, 4, 1, 5, 5
4, 1, 1
Hence the L.C.M. is 36 x 5 x 4 = 720.
Ex. 2.~'rind the L.C.M. of 15, 21, 60, 84, 140.
15, 21, 60, 84, 140
1, 7, 1, 7, 7
60
7
i; 1,
Hence the L.aM. is 60 x 7 = 420.
19. Reduction of fractions to equivalent fractions having
the least common denominator.
It is evident that the least common denominator cannot be
a less number than the L.C.M., and therefore the following
rule needs no explanation : —
Rule. — Divide the L.C.M. of the given denominators by
each denominator in turn, and multiply the cori-esponding
numerator by tb© quotient. The product thus obtained ia
30 ARITHMETIC.
the new numerator, and the L.C.M. is the least common
denominat(y*. . - • . . • i
Ex. 1. — Reduce I, -j^^, J, ^^-J to equivalent fractions, having
the least common denominator.
12
3, 12, 8, 10
1, 1,2, 5
/, Thel^C.M. isl2 x 2 )( 5 = 120.
Dividing 120 by the respective denominators, we get as
quotients 40, 10, 15, 12; and hence the given fractions
become
9 X40 ^xi 7X1 S 1 1 Xi g .
X2o' 190* iao> lao '
Or, ^, -,»,%, m> Hh
Ex. 2. Beduce to their least common denominator the fol-
lowing :—H, it, 11, 1 1-
45 30, 18 , 45, 63 Hence the L.C.M, is 45 x 2 x 7
2 ~2,"T,~I7~7 = 630.
r. 1, 7
Dividing 630 by the respective denominators, we get for
quotients 21, 35, 14, 10. Hence the required fractions ai*e
11 X a 1 XULJ.^' 69-^-lA. 8Q X 10. a.aj 4.6_5, '28, SAP.
630 . 680 630 . 630 > 630 63 630 63
Note 1. — The operation of dividing the L.C.M. of the denominators
is often simplified oy usinff the L.C.M. in its factorial form. Thus,
in our present e)cample, uie L.C.M. is 45 x 2 x 7. Now it is easy
to see that the quotient of this b^ 30 or 3 x 2 x 5 is 3 x 7 or 21, and
that the quotient by 18 or 9 x 2 is 5 x 7 or 35, and so on. We thus
avoid the process of long division.
Note 2. — It is sometimes necessary, and generally advisable, espe-
cially for beginners, to reduce the siven fractions to their lowest
terms before applying the rule for the least common denominator.
Thus, the least common denominator of the fractions f , y^, ^, taken
as they are, is 60 ; whereas, if we reduce the second fraction to its
lowest terms by striking out the factor 3, common to both numerator
and denominator, the fractions become f , f , •^, and the least common
denominator is 20. If, however, the denominator of any such frac-
tion not in its lowest terms is contained in the L.C.M. of the deno-
minators, when the fractions are all in their lowest terms, it is unne-
cessary to reduce the fraction to its lowest terms.
We strongly recommend the beginner, however, to always com*
meaoe by r&ducing the given fractiond to theix \o^^ \atiqa.
ADDITION OP FRACTIONS. 31
Ex. rv.
1. Find the L.C.M. of—
(1.) 2, 6, 8, 12. (2.) 4, 9, 10, U. (3.) 15, 21, 40,
45. (4.) 12, 20, 35, 126. (5.) 39, 65, 52, 140.
(6.) 37, 60, 222, 225.
2. Reduce to their least common denominator —
(1.) h h tV, «. (2.) *, A, H, «. (3.) I, », ih
if (4.) h A, h (5.) h A> 1%, ««.
(6.) xVff, tV^> i/W.
3. A can run round a ring in three minutes, B in four
minutes, and C in six minutes, and they start together. In
how many minutes will they all be again at the starting
point?
4. Arrange the fractions -j^, ^^, ^{9 H> ^ order of mag-
nitude.
5. Multiply the greatest of the fractions V> •???> iH ^7
339.
6. Divide the least of the fractions ,\, t^, ^^, by 6.
7. Reduce to a simple fraction the complex fraction having
the greater of the fractions -f, f in the numerator, and the
less in the denominator.
8. Which of the fractions J, ^V ^ nearer to -J 1
9. Show that ^ is less than -^.'
. * ^*
10. Arrange in order of magnitude the following :—
4f of j- lSr_ofi H
6i ' liof2V 13|ofi'
11. Show that nine times the less of the fractions tl, _._»
is eight times the greater.
12. Show that the ratio 18 i 7 is a ratio of greater in-
equality than the ratio 41 : 16.
Addition of Fractions.
20. We have shown (Art. 6) that the numerator &iidd!^TL<^
minator respectively represent the antecedent and conafco^'eiTA*
32 ARITHMETIC.
of a ratio ; and it is evident from the definition of a ratio
(Art. 6) that the sum of two ratios having the same conse-
quent is equal to a ratio whose antecedent is the sum of the
given antecedents, and whose consequent is unaltered. Hence
we have the following rule for addition of fractions —
KuL^. — Bring the given fractions to their least common
denominator, add together the numerators thus obtained, and
place under the sum the least common denominator.
Ex. 1.— Add together |, H> i\> f •
The least common denominator is easily found to be 42.
Dividing this by each of the given denominators, we get
as quotients, 14, 2, 3, 7.
Hence, ^■\.^ + ^ + ^= i2i±A + i_ox_8 + £2^.1 + Aiu
' ^ Sl'TT'6 42^42 42 48
Ex. 2.— Add together 3^, 2j, 1 JJ, 4^.
The sum of the integral parts of the given fractions =
3 + 2 + 1 +4 = 10. Hence, 3j + 2j + 1 J J + 4^v = 10 +
i + 3- + ii + tV
The least common denominator is easily found to be 180.
Dividing this by each of the given denominators we get as
quotients 36, 20, 6, 15.
Hence the required sum
= 10 + ^^3Q + 7^80 4- 11^0 + I.*^J*
180 180 180 180
- 10 + ^Vir + i*8- + 1%% + iH
= 10 + > 64-1 40 + 6 04-1 6 = 10 + 4#X
180 * ""
= 10 + mi = iiisj.
Subtraction of Fractions.
21. After the preceding article thei-e will be no difficulty
in comprehending the following rule —
Rule. — Bring the given fractions to their least common
denominator, subtract the numerators thus obtained, and
under the difference place the least common denominator.
Ex. 1. — Subtract | from J.
^-,-1— 7X5_ 3XR — 1A — 5.< — 3 5 — 24 — 11
^ ^ io' To' "" *^ ^^ ~ To ^"'
SUBTRACTION OF FRACTIONS. 33
Ex. 2.— Take 6j from 9 J.
9i-6j = 3i-j = 3 + 3^-i^
= 3 + A -« = 3+«^.
Here tlie number to be subtracted is the greater. "We
shall; however, take the less from the greater, and put the
negative sign to the remainder, meaning by this that the
remainder has yet to be subtracted.
Hence, 9^ - 6j = 3 - ii = 2 + (1 - \\)] now \\
taken from unity or ^-f evidently leaves t'^.
.'. required result = 2 + -^'^ = 2^,
We will now give an example involving both addition and
subtraction.
Ex. 3.— Find the value of Gj - 7f + 4x\ - V^.
Whenever we have an expression involving both + and
- signs, the simplest method is to add together all the
quantities affected with the pltcs sign, and likewise those
ejected with the minits sign; then taking the difference
between these two sums, we place the sign of the greater sum
before the result.
^ Thus, taking first the integers, we have 6-7+4-1 =
10 — 8 = 2 (it must be remembered that the sign + is
understood before a number which appears without a sign
when it stands alone or at the head of an expression).
Hence the given expression = 2 + ^ - f + 1^^ - tV
= 2 + ixi8 « 2x4 ^ fiX3 _ 6X2 = 2 + la-B + i s-i o
36 36 36 36 36
We shall give one more example in order to show how
brackets are to be treated.
Ex. 4.— Find the value of 7^ - (2^ - 3^) + (4^ - Ij)
The general rule, which the student will better understand
when we come to Algebra, is this —
When a minus sign stands before a bracket, it changes all
the signs within on removing the bracket ; but when a plies sign
stands before the bracket, the latter may be remo^ei mtlwal
chfW£;ing any of the signs within.
B c
34 ABITHMETIC.
Thus, taking the expression - (2^ - 3^) : —
We must remember that the sign of 2^ when within the
bracket is, according to a remark made in Ex. 3, v/nderatood
to be pltia. In fact, though it is unusual, we might write the
expression thus : —( + 2^ — 3 J).
Now, remove the bracket, and it becomes - 2^ + 3^.
Again + m - U) = + 4i - Ij.
And - (2^^^ + 3tV) = - 2^3^ - 3tV
Hence our given expression —
^ 7j - 2i + 3i + 4i - U - 2^?^ - 3^^
= 6 + 1^15 _ 1X24 + 1X6 ^ 1X2 _ 1X40 __ flXl a _^ 1X10
190 120 120 120 120 120 18 o"
= 6 + lS'^24't-fl0^-20 — 40 — 36— 10 _ g ^ 9 5 — 1 1 O
120 120
^6 -T^/^ = 6 -^ = 5 + (1 -i) = 5|.
Ex. V.
1. Add together (1.) hhil (2.) h t'^, H \ (3.) h i^ir, tV
2. Find the sum of 3|, Vl\\, 2^3^, 1,.^, |, 3|.
3. Add together 2j, ^, ^, 2j of \\, 5i of 1^^.
4. Find the difference between (1.) \ and | \ (2.) f and |;
(3.) f and ^.
5. Subtract (1.) 6 J from Sy'^ ; (2.) 3j from 4^^ ; (3.)
2^3^ from 6j.
6. TakeL;-!! from -1^,
1* 2 - li
7. Find the value of 1/^ - 2f + 3|- - \\.
8. Simplify the expression (2^ - \\) - (3f - 7|).
9. By how much does 3-,^ - 1| exceed 2\% - ^'^1
10. Take the difference of 6^^ and l^ij- fix)m their sum.
11. Add the difference of the same two fractions to their
sum.
12. Find the value of the expression ^t^ii- (| ȣf\
MULTIPLICATION AND DIVISION OF rRACTIONS. 35
Multiplication of Fractions.
22. Rule. — Multiply together the numerators of the fiuc-
tions for a new numerator, and the denominators for a new
denominator.
The reason of this rule is easily seen. Let it be required
to find the product of -J and J, or the value of | x J.
Now, what is the meaning of multiplying the ratio 3 : 5 by
the ratio 7 ; 8 ] It means evidently that the ratio 3 : 5 is to
be multiplied by 7, and the result divided by 8.
Now (Art. 8) the ratio 3 : 5, when multiplied by 7, be-
comes 3 X 7:5; and (Art. 9) the ratio 3x7:5, when
divided by 8, becomes 3x7:5x8; and we have
* =< J = i-x-^;
And so on for any number of fractions. Hence the above
rule.
Ex 1. — Multiply together the fractions -J, |, ^.
■^ >< * X H = J^t^f
Before actually performing the operation of multiplication,
it is advisable to strike out any factor common to both
numerator and denominator. We see that 5 is common to 5
and 25, 3 common to 3 and 6, 4 common to 12 and 8, and we
then have
i X f >< « = li^ = A-
The whole operation is sometimes written thus —
I >< * '< H = i^ = ^^-
2 9 6
Ex. 2.— Multiply together 2|, 3if, l^V, H-
2| X 3if X 1^ X H = V >< W X H ^ H
St<» 8 S
= 1 8X1 SMX ai X8T = A r=9,
&xsaXsGXsi A
Division of Fractions.
23. Rule. — Invert the divisor, and proceed as in multi-
plication.
To explain this rule^ Jet us endeavour to 3iv\<3L<b \\xy V
y^ema^ evidently consider <;ie required cmotleTit aa TiQ^ilt^'^
36 ARITHMETIC.
else than the ratio i : f, and this, by the reasoning of Art. 11,
is equivalent to the ratio 7x8:9x5, and we hence get
Now, f is the divisor f when inverted, and hence the
above rule.
Ex. l.-^Divide ^j^ ^7 I-
S 4
— WX» _ 18 — 9.9.
T'5^i = Axf=i^=V =
'S-
Ex. 2.— Divide If of 7^ by 3 J of 3 J.
If of 7i ^ 3i of 38- = I X V -s- (y X V)
= J X V X Vs X A = f H = ItVt-
8
We have introduced a bracket on the right side of the
first equality, for otherwise the sign - affects only the first
fraction >^. On the other side a bracket is unnecessary, for
the sign -f standing before a compound fraction (not two
fractions) affects the whole.
Ex. 3. — Simplify the expression
1t\ -^ 6| X 4J ^2f of24T^f.
The given expression = i| -r- V ^ V "^ (S ^ W)
= i J X A ^ V >< I X iW = tIt.
Ex. VI.
1. Find the sum, difference, and pixnluct of 2^ and If.
2. Multiply the sum of the fractions 3|, 2^^;^ by their
difference.
3. Simplify the expression |(3xV)*- (U^/} ^ {3^^ - U^}.
4. Beduce to a simple fraction each of the following ex-
pressions —
(1.) 1t\ -^ 7i X 8f -f 2f of 20i.
(2.) 1t-\ ^ 7i of 8 J ^ 2t -r 20i.
6. What i» the difference between (8^ - 3j) and (5 J - 4^'^)?
6, Divid. _i^ ly iil'U,
HEDtrCTION OF FRACTIONS TO DECIMALS. 37
1
7. Beduce to a simple fraction 3 +
7 + -i'V
8. Simplify the expressions (1.) i — i^ ^ ^5 — 15.
i-i H + i
& —
9. Fmd the quotient of 103 Jg by SOj of ^g j.
10. The cost of 7f articles is £65 1, what is the cost of
each article 1
1 1. Find the cost of 89 J^ articles, when one cost £^^.
12. The sum of two quantities is 34y\, and their differeneo
is ^W j required the greater.
Beduction of Fractions to Decimals.
34. If we place a decimal point to the right of an integer,
and add as many ciphers as we please, it is clear, from Art 1,
that we do not alter its value. And hence a given ratio, as
3 : 8, is not altered in value by writing it 3*000 : 8 ; and
further, dividing each of its terms by 8, according to the rule
for division of decimals, it becomes '375 : 1. It therefore
follows, putting each of these ratios in a fractional form, that
3l = 3io.p_o = -375 = '375.
® 8 1
We get, therefore, the following rule : —
KuLE. — Place a decimal point to the right of the numera-
tor, and add as many ciphers as may be thought necessary.
Divide the new numerator by the given denominator, accord-
ing to the rule for division of decimals, and, if necessary, add
ciphers to the successive remainders until the division ter-
minates, or until we have obtained as many dectcaoX ^^ic^
sjsfjvgaired
5d arithmetic!.
Ex. 1. — Keduce ^\- to a decimal.
32)5-0(-15625
32
180
160
200
192
80
64
160
160 Hence VV
Ex. 2.-
—Reduce -r^TT to a decimal
296)5-00(-01689i
296
2040
1776
2640
2368
2720
2664
560
296
264
It will be seen that vre liave arrived at a remainder, 264,
exactly the same as the second remainder ; and that, there-
fore, the quotient figures 891 will continually repeat, and
that the division will never terminate* We call 891 the
recurring period ot the decimal, and it is usual to indicate
the fact of its recurrence by placing dots over its first and
last figures, as above.
We have, therefore, as a result, ^^^^ = •016891.
Note. — It is easy to see that no fraction, reduced to its hxjoegt terms,
whose denominator contains any prime factor, other than 2 or 5, can
be expressed as a terminating decunal. For every terminating deci-
mal is an exact number of tenths, hundredths, &c., and may, there*
fore, be transformed into a fraction, having some power of 10 as its
KEDUCTION OP DECIMALS TO FRACTIONS. 30
denominator. Kow, if we wish to bring a fraction already in its
lowest terms to an equivalent fraction having some power of 10 for
its denominator, it can only be done by mulUplying its numerator and
denominator by some integer ; and it is impossible to^ obtain ajiy
power of 10 by multiplication only, from a number which contains
any prime factor, other than 2 or 5.
Redaction of Terminating Decimals to Fractions.
25. Remembering (Art. 1) that any given terminating
decimal may be considered as derived from an integer by
diminishing it 10, or 10*, 10', &c. fold, we have
•345 = 345 -r 10*.
Hence, '345 is the value of the ratio 345 : 10" j and we
have, also. *345 = = .
' ' JQ3 1000
The following rule is, therefore, clear : —
Rule. — Make a nimierator of the integer, formed by tak-
ing away the decimal point ; and for a denominator put 1, fol-
lowed by as many ciphers as there are given decimal figures*
Ex. 1.— -625 = tWtt = |-^iff = *.
3-95 = 3^*'^'^ = 3i^J- = 3^8.
•0025 = T^^^ = -^3^^ = ^i^.
Beduction of Becurring or Circulating Decimals to
lotions.
26. There are two kinds, and it is convenient to treat
them separately.
(1.) Pure cvrctdating dedmah, where the whole of the
decimal figures recur.
Rule. — Take away the decimal point and the dots, make
a numerator of the integer thus obtained, and place under
this as denominator as many nines as there are recurring
figures. •
The following example will make this rule clear.
• • •
Ex. — ^Reduce "207 to a fraction.
The value of the decimals is evidently not altered by
miting it '207207. Let us remove the decimal point three
40 ARITHMETIC.
places to the right, or, what is the same thing (Art. 1), let
us multiply the given decimal by 1000,
We then get 207*207 as the value of 1000 times the given
decimal. Now the number 207*207 includes the integer 207,
and the given decimal -207; and it therefore follows that the
integral part 207 is (1000 - 1), or 999 times the value of
the given decimal.
Hence, dividing it by 999, we get
•207 = 207 -f 999 = f JJ.
(2.) Mixed circulating decimals. — ^Where part only of the
figures recurs.
BuLE. — ^Take away the decimal point and the dots, sub-
tract from the integer thus obtained the integer formed
by the figures which do not recur, and make a numerator of
the result. Then, for a denominator, place underneath as
many rdnes as there are recurring figures, followed bv as
many ciphers as there are figures which do not recur.
We shall make this rule clear by the following example : —
Ex.^ — ^Keduce '24573 to a fraction.
Let us remove the decimal point two places to the right,,
thus, by Art, 1, multiplying the given decimal by 100 ; wa
then get
100 times the value of -24573 = 24-573.
Now, by case (1) above, 24-573 = 24j^f ; or reducing
to an improper fraction, and noticing that 24 x 999 =•
24000 - 24, we have
24*573 = 24000 — 24 + 573 — 24578 -• 24
9 9 9 9 9 9
Hence, dividing this result by 100, we get
•24573 = 24573-a^ ^ 100 = g^^-^^-g-* (Art. 24.)
999 99900 y-^-" »" •/
Ex. 1.— -428571 = HUH = ?-$-^-^^^^' = f
» V » V V V 7X142867
■Ry 9 •97nQ — 2 7 09 -• 27 ' — 5502. — 140x2x9 4.4fi.,
±iX. J.— J/Uy ggoO ~ »»^« ~ 660 X 3 X 9 **y
27. In arithmetical operations involving circulating deci-
mals, and, indeed, any decimals having a large number of
decimal figures, it is generally sufficient to obtain a result
correct to a given number of decimals.
APPBOtlMATES RESULTS. 41
1. In addition and siihtraction we obtain this result most
easily by using in our operation one or tif)o more figures of
the given decimals than are required in the result.
Ex. 1. — ^Add together (correct to five places) the follow-
ing:—
•3026, 6-7294, -016, -4163729.
•3026026
6-7294444
•0166666
•4163729
7-4650865. Ans. 7-46508.
Note. — If onr object is to obtain a decimal of five places which
shall give the approximate sum of the given decimals we must write
7*46509 as the approximate sum; for 7 '46509 is nearer to the true
value of 7-465086 oc. than 7-46508. The general rule is to increase
by 1 the decimal figure at which we stop, wJien the next figure is 6 or
above 6,
Ex. 2. — Find the difference (correct to six places) of 3*0745
and 4 '263, and express the approximate difference by a deci-
mal of five places.
4-26326326
3-07454545
1-18871781
Hence the difference correct to six plaqps is 1*188717, and
the required approximate difference 1-18872.
2. In multiplication and division of circulating decimals it
is generally preferable to reduce the given decimals to frac-
tions, bring out the result in a fractional form, and afterwards
reduce this to a decimal.
Ex. VII.
1. Express as a decimal the sum of the following fractions: —
2. Reduce to fractions the following decimals : —
•35, -026, -16, -142857, -16, -4285714.
• • • •
3. Find the value (correct to six places) of -237 + 3*816 —
6-0235 + 4-29 - -002 + 1*374.
42 ARITHMETIC.
4. Add together -62, -037, 2-476i, -SiOG, -7, -375.
5. Find the product and quotient of 3*6 4 by 4*3.
6. Simplify the expression —
(4-6 X -428571 - 2-2 x -36) (1 - -16).
In the next six examples the dots are signs of multiplica-
tion, and you are required to give the values of the expros-
sions correct to six places.
^' V2 ~ rsl ^ rM4 " 1-2-3-4-5 * *°'
irt 1 1 1 1 ^
10. 1 + gj- + ^ + gj + &c.
^^- ^^{l - 3T» "-si - W -" ^i " *{ 23
3-239*
+ (fee.
10 1 1 1 ^ 1 4. -L
^^- r2 "*■ ITS *** 34 "*" f5 *^ 5-6 ■** 6-7
CHAPTER IH;
APPLICATION OP THE PRECEDING ARTICLES TO CONCRETE
QUANTITIES.
To find the Value of a Fraction of a Concrete Quantity.
28. Rule. — Multiply the concrete quantity by the nu-
merator of the fraction, and divide the product by the deno-
minator.
VALUE OP A FEACTION OP A CONCRETE QUANTITY. 43
Ex. 1.— Find the value of f of 2 tons, 3cwt. 21 lbs.
f of 2 t0nS,3 CWt. 21 lbs. - <'*°'^°°^/^""^^' = <ton,.9ewt8q>.7lb,.
= 18 cwt. 2 qrs. 1 lb. Ans.
Ex. 2.— Find the value of 3j of X12. 6s. 2Jd.
£. 8, d,
3timesX12. 6s. 2id. = (XI2. 6s. 2Jd.) x 3 = 36 18 6^
Hence, adding, the value required =X39 13 3j^
Ex. 3.— Find the value of f of 4 miles + J of 3 fur. + ,«r
of 8 poles.
Iliile. Fur. Poles. Tarda.
f of 4 miles = '*-^^|2^ = ^ = 1 5 28 3^
i of 3 fur. = a^^ = «i!- = 2 13 . If
A of 8 poles = ^^^^ = *!^ = 1 4j^
Hence, adding, the value required =203 i\%
Note. — ^The addition of the yards is thus effected : —
(af + H + -HJ) yds. = (8 + ^^^) yds. = (8 + m) yds.
= W yds. = 1 pole,(3i + I J) yds.
= 1 pole,(3 + 2^^) yds. = 1 pole,4H yds.
Ex. VIII.
Find the values of —
1. I of XI; I of Is.; ^ of 12s.; | of £3.
2. J of X5; f of a guinea; -^ of 2s. 6d.; f of a crown.
3. 3f of £1. 12s.; 2^ of X3. 10s.; 7^ of £3. 4s. 5 J.
4. 4 0^ 1 ton ; I of 1 qr. ; J of 1 stone ; -^ oi^ lbs.
5. 3f mile; J of 3 fur.; f of 15 poles; 3| of 2j of 12 yds.
6. oTTT leap year; -rg luiiar month; yTjj o^ 1^ ^« 15' 12^.
7. 12 lbs. 3 oz. 7 dwt. 5 grs, x 3^; 10 oz. 4 gr. -r 16^.
44 AttlTHMETlC.
8. 8fof4ac. 3po. - ^V of 1 sq. mile + If of 3 r. 20 sq. yds.
9. m of A -^ ItV) of 35« 36' 25^; (2^^ 4- Ij of 1 J) of 30^
10. T?Ty of 1 lb. Troy + -J of 1 lb. Troy-^V of 1 lb. Avoir-
dupois ( = 7000 grains).
12. 2^3^ of 15 h. 10' 13i^ - /^ of 1 day, 12 h. 11' 12^^
To Bednce a given Quantity to the Fraction of any
other given Quantity of the same kind.
29« BuLE. — Keduce both the given quantities to the same
denomination, and the fraction required will have the number
of units in the first quantity as numerator, and that of the
Becond quantity as denominator.
Ex. 1. — Reduce 3s. Sd. to the fraction of £1.
3s. Sd. = 44d,
and^l = 240d.
Hence, the fraction required = ^^j^ = ^J.
Or, better, thus :
3s. 8d. = 11 fourpences,
and £1 = 60 fourpences.
.*. Fraction required = -J^.
KoTE. — It is always best to keep the denominations to whicb tlie
given quantities are reduced as high as possible.
Ex. 2. — Reduce ^ of a moidore to the fraction of 2 J guineas.
I of a moidore = (| x 27)s., and 2J guineas = (2J x 21)8.
Hence, the fraction required = = - — ~ = 4-x H x J
^ 2^x21 Jx21 *^«>^'
_ 1 xp xa _ 1 B
6X7X7 246
Ex. 3. — Reduce 3 cwt. 2j^ qrs. to the fraction of 4 cwt
2 qra 4 lbs.
3 cwt 2j^ qrs. = 14j^ qrs., and 4 cwt. 2 qrs. 4 lbs. = 4 cwt
2| qrs. = 18| qrs.
14i -12-i
.-. Fra^jtion required = jg» = jfl = JVlTV = 2-
' 7
VALUiS OP A DECIMAL OP A CONCRETE QUANTITY. 45
Ex. IX.
Beduce —
1. Is. 8d. to the fraction of ^1 ; Tjd. to the fraction of 10s.
2. 2s. 4d. to the fraction of 10s. 8d.; Is. 7^d. to the frac-
tion of 3s. 4^d.
3. 3 qrs. 15 lbs. to the fmction of 1 ton; 2 stones 10 oz.
to the fiction of 3 cwt.
4. 3 lbs. avoirdupois to troy weight; 10 lbs. 3 oz. 4 dMrt.
troy to avoirdupois.
5. 3 quires, 10 sheets to the fraction of 2 reams, 3 quires ;
3 ft. 8| in. to the fraction of 3 yards.
6. 30^ 3' 12* to the fraction of a right angle ( = 90«^);
57® 16' 21^'V'' to the fraction of two right angles.
What fraction is —
7. f ac. of 3^ ac.; 2 J days of 17 weeks 1
„ •125 + 1-875 ,,, , 3-16 X 1-4 ^ ^^ ^
^' '^40625' ^^^^ ^^ ^^* P^^^« > 2.375x1 ^^' ^^ ^^ «^ 1
9. ( j7 rood + jz poles + — -^ yd. ) of 3 acres i
10. o-^ gaL of ^^ pipes 1
11. What fraction of his original income has a person left
after paying a tax of 4d. in the £ ?
12. A garden roller is 2 ft. 6 in. wide, and it is rolled at
the rate of 1 mile in 20 minutes : find in what fraction of a
day a man will roll } of an acre.
To Find the Value of a Decimal of a Concrete Quantity.
SO. Rule. — Multiply the given decimal by the number of
units in the concrete quantity when expressed in terms of one
denomination, and the integral part of the result will be the
number of units of this denomination. Then multiply the
decimal part of this denomination by the number of units
connecting it with the next lower and the integral part
yfiH b^ vuiits of this fattier denomination^ and so on,
46 ARITHMETIC.
Ex. 1.— Find the value of -325 of £3. 10s.
£3. 10s, = 70s.; proceeding then according to rule, we have:
•325
70
22-7508.
12^
900d. Ans.: 22s. 9d. or XI. 2s. 9d,
Ex. 2.--Find the value of -546875 of 3 tons.
•546875
3
1-640625 tons.
20
12-812500 cwt.
4
3-2500 qrs.
28
200
50
7-00 lbs. Ans.: 1 ton, 12 cwt. 3 qrs. 7 lbs,
Ex. 3.— Find the value of 6-66875 acres,*
6-66875 acres.
2-675.00 roods.
40
27-000 poles. Ans. : 6 acres, 2 roods, 27 poles,
Ex. 4.— Find the value of -316 of XI.
First Method. Second Method.
£, .31g _ 316 - 31 ._ 3BJt
•3166667 nearly. ^ ,/^'°, _ J'"
20 "" 16 X 60 ■" 60
6-3333340s. Then by rule for reduction
X2 of fractions,
4-000008d. Ans.: 6s. 4d. Xj?. = 6s. 4d. Ans.
60
The latter method is preferable when perfect accuracy is
required.
VALUE OF A DECIMAIi OF A CONCRETE QUANTITY. 47
Ex, X.
Eind the value of —
1. £-375; £-98125; £-815625.
2. -416 of £3; -428671 of 6s. 5d.; 8*671428 of 3s. 0|d.
3. -625 of 1 ton; -046876 of 3 tons; 4*39 of 1 cwt. 53 lbs.
4. 1-6671875 acres; -3475 rood; -076923 of 5 acres, 6poles.
5. -2083 of 1 ream; -4583 quire; -383 of 3 reams, 12 sheets.
6. -3078125 pipe; -490626 tun; -37125 bushel
7. Express in grains -142857 of 4 lbs. avoirdupois, and
express ihe result in troy weight.
8. Express 10 oz..3 dwt. 14 grs. as the decimal of 1 lb.
avoirdupois.
9. What is the sum of -6 of 1 guinea, -083 of 1 crown,
and -037 of £1. Os. 3d.
10. Find the value of
(-20416 X 7-5\
4 ^ .^^gyg — I ewt. + (-95 ton- 769230 qrs.) x 26.
11. What is the value of
(3 + — L- ) of 1 cwt. 35 lbs. 1
1 + :
•3
12. Simplify / l±J:f + ^—^\ of 1 oz. 15 dwts.
\1 - -16 1 + -16/
To Beduce from one Denomination to the Decimal of
another Denomination of the same kind.
31. Rule. — ^Bring the given quantity to the fraction of
the proposed denomination^ and reduce thia fra^tioiv, \a ^
48
ABITHHETIC.
Ex. 1. — ^Reduce 3s. 3d. to the decimal of 8s. l^d.
3s. 3d. = 26 three-halfpence,
8s. l^d. = 65 three-halfpence,
and the fraction H = «->Lj_s = | = -4.
°* 6X13
Hence -4 is the decimal required.
Ex. 2. — Reduce 3 qrs. 21 lbs. to the decimal of 2 cwt. 3 qrs.
3 qrs. 21 lbs. = 105 lbs.
2 cwt. 3 qrs. = 308 lbs.
and the fraction ig4 = i-&-^' = ii = -^^^ = '3409.
308 44X7 ** 11 ,
Hence *3409 is the decimal required.
Ex. 3. — Reduce 18s. 8^ to the decimal of .£1.
The rule may be applied most concisely as follows : —
4
12
20
8-25
18-6875
•934375 .-. -934375 is the decimal required.
The farthing is first reduced to the decimal of a penny, and
the 8d. prefiLxed; then 8*25d. are reduced to the decimal of a
shilling, and the 18s. prefixed; lastly, 18 '6875s. are reduced
to the decimal of £1.
Ex. 4. — ^Reduce 3 qrs. 21 lbs. to the decimal of 1 ton.
4
20
21
3-
3-75
•9375
•046875 Hence -046875 is the decimal required.
Ex. XI.
Reduce —
1. 12s. 6d., 10s. 7td., lis. l^d., 18s. 6}d., each to the
decimal of £1.
2. 13s. 0}d., 10s. 8jd., 9s. 6d., each to the decimal of
158. 5^d.
3. 7a 8f d. to the decimal of a guinea ; and 3s, 2^d. to the
decimal of f^ moidore,
THE METBIO SYSTEM. id
4. 1 qr. 7 lbs. to the decimal of 1 ton ; 3 cwi 3 qrs. 20 lbs.
to the decimal of 3 tons.
5. 3^ lbs. to the decimal of lu cwt. ; 14 oz. to the decimal
of 3 cwt. 2 qrs.
6. 20 grs. to the decimal of 1 lb. Troy; 3 dwts. IG grs.
to the decimal of 4 oz. 11 dwts.
7. 1 rood, 10 poles to the decimal of 1 acre ; 3 roods, 15j-
square yarda to tiie decimal of 5 acres.
8. f hours to the decimal of 10 weeks; 7 h. 18' to tlie
decimal of 1 year (365 days).
9. Bring the sum of ^V ^^ 9 hours, | of 12j^ days, J of 7 J^
minutes, to the decimal of a week.
10. Express a pound troy as the decimal of a pound avoir-
dupois.
11. Reduce the sum of 6 lbs. 6f oz. avoirdupois, and
8 oz. 6 dwts. 16 grs. to the decimal of 1 ton.
1 2. Express 2-36 of 4s. - '51 8 of 9s. 2d. + 1-4583 of 6d. as
the decimal of £5.
CHAPTER IV.
THE METRIC SYSTEM.
32. — ^The fundamental unit of the metric system is the
metre. A metre is the ten-millionth, or yo? V^^ of 90° of
the earth's meridian, and measures 39*3708 English inches.
In order to express multiples and sub-multiples of this unit,
and, indeed, of any unit in the metric system, we make
use of one or more of the following prefixes : —
Beka, . . 10 times. Deci, . . 10th.
Hecto, . 100 „ Centi, lOOth.
Kilo, . . 1,000 „ Milli, . . 1,000th.
Myria, . 10,000 „
We will arrange these prefixes and the word nnit in order
according to their signification, thus —
Myriaj KUo, Hecto, Deha, Unit, Deci, Centi, Milli,
If ow, as wo yefi4 this Ji»« from }eft t)o right, it iq evident
6 J)
50
ARITHMETIC.
that tlie words have a Bignification decreasing tenfold invalae;
and as we read it from right to left, they have a signification
increasing tenfold in value.
It therefore follows that figures placed under the ahove
words have a local as well as an intrinsic value ; and further,
if when a figure is wanting to complete the series, its place
be filled up by a cipher, it will be seen that the local value
corresponds exactly with the ordinary decimal notation.
Moreover, we have only to place a mark (in fact, a
decimal point) at the right of the figure standing imder any
of the words of the above memonal line, and the given
quantity is at once expressed in the denomination corres-
ponding to that figure.
Thus, taking the metre as our unit :
M>Tiam. Kilom. Dekam. Metres. Decim. Millim.
3 2 5 4 7 8
Hyriam. Kilom. Hectom. Dekam. Metre. Decim. Centim. Milllm.
= 32064708
= 3-2054708 myriametres.
r^ 32-054708 kilometres.
= 320-54708 hectometres.
= 3205-4708 dekametres.
= 32054-708 metres.
= 320547-08 decimetres.
= 3205470-8 centimetres.
= 32054708- or 32054708 millimeti^s.
The following rule for expressing any quantity in terms of
any one multiple or sub-multiple of the unit, or of the unit
itself, is therefore evident : —
Rule. — Put ciphers in the place of any multiple, unit, or
sub-multiple absent in the series, and write the figures in close
order, as in the ordinary decimal notation. Then place a
decimal point at the right of the figure corresponding to the
denomination in which we wish to express the given quantity.
Ex. 1. — Express 5 myriam. 3 hectom. 6 decim. as metres.
Filling up with ciphers the vacant spaces, we have —
Myriam. Kilom. Hectom. Dekam. Metres. Decim.
5 3 6
= 50300-6 tne^jtea.
THE METRIC SYSTEM, 51
Ex. 2. — Express 3 dekam. 4 decim. as myriametres.
Filling up with ciphers, we have —
Hyriam. Kilom. Hectoxn. Dekam. Metres. Decim. Centim. Millua
00030400
c 0'0030400 jnyriametres = 0*00304 myriametres.
(The student will see thiat it was unnecessary here to
extend the series beyond decimetres.)
Ex, 3. — Express 13 metres 502 millimetres as kilometres.
We may write the given quantity thus —
Kilom. Hectom. Deckam. Metres. Decim. Centim. Millim.
13 5 2
= 0-013502 kilometres.
We have hitherto spoken only of the fundamental unit, and
its multiples and sub-mviltiples. We shall hereafter (Art. 35)
explain ilie principal derived units, viz», the Gram, the Are,
the Stere, the Litre, and the Franc ; but as the multiples and
sub-multiples of these derived units bear the same relation
respectively to the corresponding derived unit, as in the case
of the fundamental unit, all the preceding remarks relative
to the multiples and sub-multiples of the fundamental unit
apply equally to those of the Gram, the Are, the Stere, the
litre, and the Franc.
With regard to the units, multiples, and sub-multiples of
square and cubic measure, properly so called, it is necessary
to make a few remarks.
SS. Square Measure. — The unit of square measure is the
sqttare metre; and since the series myriam., Idiom,, hectom.,
dekam,, metre,, (kc., decrease in value tenfold when read from
left to right, and increases similarly when read from right
to left, it follows that the series square myriam., square
kilom., square hectom., square dekam., square metre, &c.,
will decrease or increase 10^ or 100-fold. Hence we see that
in square measure the multiples and sub-multiples increase
or decrease successively 100-fold, and, therefore, when quan-
tities in square measure are expressed by the ordinary
decimal notation, each multiple or sub-multiple must occupy
the place of two figures, a cipher being supplied when.^^
have less than ten of any multiple or sub-multiple, soA X*"^^
et/f£eivw£eD there 18 anjr blank in iheB&A'&fis —
52 ABITHMETIC.
Sq. kilom. Sq. hectom. Sc^. dekam. Sq. metre. Sq. centim.
Ex. 10 3 15 3 5
8q. kilom. Sq. hectom. Sq. dekam. Sq. metre. Sq. decim. Sq. centim.
z^ 10 03 15 03 00 05
^ 10-0315030005 square kilometres.
= 1003-15030005. „ Lectometres.
= 100316-030005 „ dekametres.
=: 10031603O005 „ metres.
=: 1003150300-05 „ decimetres.
= 100316030005 „ centimetres.
34. Oubio Heasnre. — The unit of cubic measure is the
mjihic metre, and hence after the remarks in the last article,
since 10* = 1000, when quantities in cubic measure are ex-
pressed by the ordinary decimal notation, the units, mul-
tiples, and sub-multiples must respectively occupy the place
of ihree figures, ciphers being supplied to fill up blank
spaces when necessary.
Ex. 1. 326 cubic metres 51 cubic decimetres.
= 325 „ „ 051 „
= 325*051 cubic metres,
s 325051 cubic decimetres.
Ex. 2. — 25 cubic metres 3 cubic decim. 40 cubic centim.
= 25 cubic metres 003 cubic decim. 040 cubic centim.
=r 25-003040 cubic metres.
= 25003 040 „ decimetres.
= 25003040 „ centimetres.
86. Derived Units. — ^The principal derived units of the
metric system are —
1. The Oranii for measures of weight.
The gfram is the weight of a cubic centimetre of distilled
water at the temperature of 4^ C.
1 gram = 15*4323 grains, 1 grain =? »Q648 gi-am.
2. The Are, tov land measure.
The (»rd is a square whogp side measure^ ^0 metres ; it is
tbertfm e<)Uftl tQ ft t%w^ Afkmetrei of IQQ ^%^9^ metres.
THE METRIC 8TSTE1C.
63
1 are = 119-6033 square yards, 1 hectare = 2*471 acres,
1 acre = 405 hectare.
3. The Stere, for fire-wood.
The stere is equivalent to a cubic metre. It is therefore
the solidity of a cube whose edge measures 1 metre.
4. The LitrOi for measures of capacity.
The litre is a capacity equal to the volume of a cube
whose edge measures a decimetre or 10 centimetres. It is
therefore equal to a cubic decimetre or 1,000 cubic centimetres,
and 1,000 litres are equivalent to a cubic metre.
1 litre = *2201 gallon, 1 gallon = 1*543 litres, 11 gallons
= 50 litres nearly.
5. The FranOy for money.
The franc is a coin weighing 5 grams, and composed of an
alloy, nine-tenths of which are silver and one-tentii copper.
The following table exhibits at a glance the fundamental
unit and the above derived units, together with the multiples
and sub-multiples at present in use : —
Table op the Metric System op Weights and Measures.
Multiples.
Units.
SUB-MULTIFLES.
10,000
1,000
100
10
METRK,
Long Measure.
10th.
100th.
1,000th.
Mjria.
Kilo.
Hecta
Deka.
DecL
CentL
MillL
Mjria.
Kilo.
Heoto.
Deka.
GRAM,
Weight.
Ded.
CentL
Milli.
Hecto.
ARE,
Land Measure.
CentL
Deka.
strrk,
Solid Measure.
DecL
Hecto.
*
Deka.
LITRE,
Capacity.
FRANC,
Money.
DecL
Decime.
CentL
Milli.
Centime.
A quintal = 100 kitog. = 2 cwt. nearly ; a millier, or tonneau de mer, s 10 quintals
s= 20 cwt. nearly.
54 ABITHMETIC.
Ex. XII.
Examples upon the Multiples and Sub-Multipled of the
Units.
1. Express each of the following as metres —
15 myriam.; 20 kilom.; 1 hectom.; 27 dekam.; 25 decim.;
100 centim.; 345 centim.; 5294 millim.
2. How many centimetres in the following —
46 myriam.; 30 kilom.; 295 hectom.; 1*5 dekam.; 3*95
metres, 295 millim. 1
3. Express according to the metric table —
20 kilog. 29 dekag.; 18 kilog. 85 decig.; 123 hectog.
13 centig.; 12 dekag. 296 millig.; 153 centig. 3 millig;
3427 millig.
4. How many decigrams in the following —
16 kilog. 12 centig; 25 hectog. 10 grams; 39,645 millig.;
20 kilog. 35 dekag. 5 grams 1
5. Express in ares —
lOhectar. ; 296 centiar.; 29 hectar. 3 centiar.; 3 hectar.
12 centiar.; 376,543 centiar.
6. How many square decimetres are there in the following —
100 sq. kilom.; 10 sq. hectom.; 5 sq. dekam.; 3498 sq.
met. ; 46 sq. met. 1
7. How many steres in the following quantities —
15 dekas. ; 394 decis. ; 9 dekas. 2 decis. ; 186 dekas. 3 decis.;
3764 decis. ; 4 decis. 1
8. Express as cubic metres —
10,000 cubic decim. ; 1,234,567 cubic centim ; 372,456,126
cubic millim.; 1,000,000 cubic centim.; 639 cubic centim.;
293 cubic decim.
9. Express as litres —
3kiloL 2 hectol. 3 litres; 4 decil.; 2 kilol. 3 millil.;
76,384 milHL; 2934 centil.; 830 dekal.; 34,576 decil.
10. Express as litres, as dekalitres, and as centilitres —
18 kiioL 3 hectol. 4 decil. 5 centil. 3 millil.
11. How many francs in the following sums of money —
100 cent.; 736 dec; 24,645 cent.; 5 cent. 25 dec;
1695 cent. 1
^E METRIC df StEll. 55
12. How many centimes and how many decimes are expressed
by the following —
13 francs; 7 fr. 13 c; 12 fr. 3 dec. 6 c.; 29 c; 3 fr. 2
dec; 18 fr. 4c.1
Addition, Subtractioni Multiplication, and Division
in the Metric System.
36. Since (Art. 33) all quantities in the metrical system
may be expressed as one denomination by figures whose local-
as iirell as intrinsic values follow the decimal system of nota-
tion, it is evident that when they are so expressed we may
add, subtract, multiply, or divide them exactly as ordinary
intc^rs and decimals.
Ex. 1. — Add together 49 metres 36 centim. ; 3 kilom. 2
dekam. 3 decim.; 2 hectom. 3 metres 25 centim.; and 13
dekam. 327 miUim.
Metres. Millimetres.
49-36 49360
3020-3 3020300
203-25 or thus, by integers— 203250
130-327 130327
3403-237 3403237
Ans. : 3 kilom. 4 hectom. 3 metres, 2 dedbn. 3 centim. 7
millim.
Ex. 2.— Subtract 3 kilog. 2 dekag. 37 millig. from 10
myriag. 25 grams, 369 millig.
Kilograms. Milligrams.
100-025369 100025369
3-020037 or thus, by integers— 302003 7
97-005332 97005332
Ans. : 9 myriag. 7 kilog. 5 grams, 3 decig. 3 oentig. 2 millig.
Ex. 3. — Multiply 12 dekasteres, 3 steres, 5 decisteres by 23.
Steres. Decisteres.
123-5 1235
23 23
3705 or thus, by integers— i3705
2470 2470
2840-5 steres. 28405 decisteres.
Ans. : 284 dekasteres, 5 decisteres.
56 ARITHMETIC.
Ex. 4. — Multiply 456,602 cubic centimetres by 36.
Cubic Metres. Cubic CentimetreB.
0-455602 455602
36 36
2733612 or thus, by integers— 2733612
1 366806 1366806
16-401672 16401672
Ans. : 16 cub. met 401 cub. decim. 672 cub. oentim.
Ex. 5.— Divide 1369 kiloL 35 Ht 36 centU. by 72.
Kilolitres. Centilitres.
136903536
^9 / ^ ^ 36^'Q3536 r
'"18 152-11504 or thus, by integers— ^^ I
9
152-11504 or thus, by integers — "" I 8
T9-01438
15211504
1901438
Ans.: 19 kilol. 1 dekal. 4 lit. 3 decil. 8 centiL, or 19 kiloL
12 lit. 38 centil.
Ex. 6. — How many times is 12 sq. dekam. 3 sq. met 15
sq. decim. contained in 216 sq. dekam. 56 sq. met 70 sq.
decim. ]
Sq. Dekam. Sq. Dekam. Sq. Decim. Sq. Decim.
120315)216-5670(18 120315)2165670(18
120315 ^ ^ . 120315
"962520 or thus, by mtegers— 952520
962520 9 62520
Ans.: 18.
Ex. XIII.
1. Add together —
(1.) 3 metres, 2 decim. 4 centim.; 18 metres, 219 millim.;
4 kilom. 2 hectom. 3 dekam. 14 centim.; 12 kilom. 36
metres.
(2.) 74006 hectom., 3216 kilom.; 12 myriam. 2167
metres.
(3.) 4 sq. met 42 sq. decim. ; 12 sq. dekam. 18 sq. decim.;
82 sq. met. 3250 sq. decim.; 3*271 sq. met.
(4.) 18 cub. met 186 cub. decim.; 39-207365 cub. met
30761 cub. centim.; 12 cub. met. 124*27 cub. decim.
THE METRIC SYSTEM. §7
(5.) 25 kilog. 235 grams; 3072 centig.; 13 kilog. 51
grams, 63 mil%. 813207 decig.
(6.) 319 hectar. 4 ares, 51 centiar.; 93-712 hectar. 2375G-27
ares ; 6 hectar. 4 centiar.
(7.) 3 steres, 5 deds.; 209 steres, 4 decis.; 25*76 stores,
13-027 dekas.
(8.) 51 kilol. 126 lit. 32 centil.; 123 lit. 3 centil. 15-02703
kilol.; 12 kilol. 3*27602 hectol.
(9.) 161 fr. 35 c; 32 fr. 4 c.j 8276 c; 10-26 fr.; 16
dedmes, 5 c.
2. Subtract —
(1.) 4 metres, 372 millim. from 16 hectom. 5-06 metres.
(2.) 30765 centim. from 12 kilom. 4 metres, 9 millim.
(3.) 3 sq. met. 89 sq. decim. from 1 sq. dekam. 7 sq. decim.
(4.) 12*0324 cub. met. from 18 cub. met. 29 cub. millim.
(5.) 39 grams, 65 millig. from 6 kilog. 12 grams.
(6.) 8 hectar. 19*08 ar. from 32 hectar. 70 ar. 2 centiar.
(7.) 9 dekas. 6 decis. from 50 dekas. 2 decis.
(8.) 6 kilol. 6 miUil. from 700 kilol. 3 lit. 3 centil.
(9.) 65 c. from 3 fr.; and 2 fr. 4 c. from 100 fr. 60 c.
3. Multiply —
(1.) 10 metres, 35 millim. by 7, 11, 13.
(2.) 18 kilom. 3-07 metres by 27, 48, 64.
(3.) 3-0625 sq. met. by 16, 18, 35.
(4.) 4 cub. met. 10 cub. decim. 5 cub. millim. by 19, 23, 26.
(5.) 7364 hectog. 9*31 decig. by 15, 25, 20.
(6.) 12 hectar. 3 centiar. by 30, 50, 40.
(7.) 416 steres, 2*9 decis. by 100, 150, 60.
(8.) 612305-06 Htres by 12, 14, 16.
(9.) 39 fr. 10 c. by 75, 105, 135.
4. A merchant owed 1500 fr., and he gave in payment 69
metres of cloth at 3 fr. 4 c. per metre, 48 metres of silk
at 8 fr. 65 c, 13*5 metres of calico at 75 c. How
much does he still owe ?
58 ARITHMETIC.
5. Make out the followiBig bill —
fr. c.
44 hectol. of oil, . . at 75 the litre.
66 kilog. 125 gr. of sugar, „ 1 25 „ kilog.
375 gr. bf pepper, . . „ 3 5 „ „
128*75 hectog. of soap, . „ 1 75 „ „
562 gr. 5 decig. of coffee, . „ 30 „ hectog.
6. Divide —
(1.) 17 metres, 16 centim. by 11, 12, 13.
(2.) 41 kilom. 82 dekam. by 15, 16, 17.
(3.) 29 sq. met., 2740 sq. centim. by 14, 21, 35.
(4.) 376-38 cub. met. by 9, 27, 45.
{5.\ 4 kilog. 14 dekag. 18 decig. by 22, 33, 55.
(6. 1 8 hectar. 58 ares by 65, 60, 55.
(7.) 12 dekas. 1-2 decis. by 12, 13, 91.
(8.) 3|6 myrial. 4 kilol. 16 lit. 7 decil. by 9, 18, 27.
(9.) 7339 fr. 50 c. b^ 25, 30, 75.
'^. Find the price of —
(1.) A metre, when 2 met. 80 centim. cost 70 fr.
(2.) A square decim., when 30 sq. met. cost 450 fr. 30 c.
(3.) A cubic metre, when 15 cub. decim. cost 361 fr.
80 c.
(4.) A hectometre, when 3 kilom. 125 metres cost 10 fr.
25 c.
(5.) A kilog. of coffee, when 7 hectog. 50 grams cost
1 fr. 35 c.
(6.) A hectare, when 4265 fr. 2*50 c. is the price of 149
ares, 25 centiar.
(7.) A stere, when 125 dekas. 4 decis. cost 20631 fr. 60 c.
/8.^ A decilitre, when 47 dekal. 6 litres cost 570 francs.
(9.) A cub. centim., when 1 cubic metre cost 10,000 fr.
8. How many times is —
(1.) 1 kilom. 470 met. 38 centim. contained in 36759-50
metresi
(2.) 12 sq. decim. 75 sq. centim. contained in 10 sq. met.
20 sq. decim. 1
(3.) 13 sq. met. 25 sq. decim. contained in 318 sq. dekam.?
THE METRIC SYSTEM. 59
(4.) 31 cub. met. 725 cub. decim. contained in 45684
cub. met. ?
(5.^ 345 millig. contained in 165 kilog. 6 bectog.l
(6.) 275 centiar. contained in 396 hectar. 1
(7.) 7 steres, 2*5 decis. contained in 29 dekas.?
(8.) 4 kilog. 5 grams, contained in 38 myiiag. 4 kilog.
480 grams ]
(9.) 8 centimes contained in 10 francs?
9. A merchant bought 95 litres of wine for 118 fr. 75 c, and
sold it at a loss of 10 c. per litre. What was the selling
price per kilolitre 1
10. To make 12 suits of clothes, it required 40 metres of
stuff 90 centim. wide. How much stuff will it take if
the width is 80 centim. 1
11. How many cubic decimetres of iron are there in a bar
weighing 280 kilog. 368 grams, when one cubic centim.
weighs 7 grams 788 millig. 1
12. An iron wire, 126 metres long, is cut into pieces 3 centim.
2*5 millim. long. How many pieces are there ]
Belation between the Metric Units and the English
System of Weights and Measures.
37. "We shall work a few examples to show how quantities
expressed in the metric system may be expressed in the
English system, and vice versa.
Ex. 1. — Reduce 10 kilom. 321 metres to English measure.
10 kilom. 321 metres = 10321 metres.
= (10321 X 1-094) yards.
_ T0321 X 1.094 -1
17 60 mues.
- 6 miles 731 -174 yards.
Ex. 2. — Express 2 miles, 309 yards in the metric system.
2 miles, 309 yards =(2 x 1760 + 309) yards.
= 3829 yards = f;^^- metres.
= 3500 metres.
= 3 kilom. 500 metres.
60 ARITHMETIC*
/
Ex. 3. — Beduce 1 ton to kilograms^ having given 1 gram
= 15-4323 grains.
1 ton = 20 X 112 X 7000 grains.
= 1016050-7507 grams nearly.
= 1016 kilog. 50 grams, 750-7 millig. nearly.
Ex. 4. — Express £13. 17s. 4^d. in the pound and mil
system.
(.£1 = 10 florins, 1 florin =10 cents, 1 cent = 10 mils.)
Reducing the given sum to the decimal of a pound, we
XlfliVG——
£13. 17s. 4|d. = X13-86875.
= £13. 8 fl. 6 cent 8f mil
Ex. XIV.
1. Express a mile in the metric system, having given that
a metre = 39*3708 inches.
2. An are contains 1076*43 square feet. Reduce 53 ares
25 centiares to English measure.
3. The area of a room is 22 sq. met. 26 square decim.
Express this in English measure (1 metre = 39*3708 inches).
4. A block of marble measures 3 feet, 3 inches in length,
2 feet, 6 inches in depth, and 3 feet, 9 inches in widtL What
is the solid content expressed in cub. centim. ?
5. In 1235 litres how many gallons, when 50 litres = 11
gallons nearly ?
6. Sujiposing a franc to be equivalent to 9Jd., reduce
£44. 13s. to francs.
7. Taking £1 sterling as equal to 25*22 francs, reduce
£2. 13s. 7id. to francs.
8. In 1852 France reaped about 47850000 hectol. of
wheat. Express this in gallons, assuming 1 gallon = 4*543
litres.
9. The ceiling of a room contains 83 sq. met. 53*96 sq.
decim. What will be the expense of painting it at lOd. a
square yard (1 metre = 1*094 yard)?
10. Find the cost of 2000 kilog. of sugar at fr. 50 c. per lb.
PROPOBTION. 61
11. In England the unit of work is the foot-pound, and in
the metric system it is the kilogram-metre. Beduce 62
metric units of work to English units, taking 1 gram ^
15-4323 grains, and 1 metre = 39*3708 inches.
12. The pressure of the atmosphere is 14 J lbs. upon the
square inch. Find the pressure in kilograms upon the squai*e
centimetre.
CHAPTER V.
FBOPOBTIOK.
38. Proportion is the equality of ratios.
Thus, since the ratio 6 : 8 = f = } = i|S,
we have ratio 6:8 = ratio 15 : 20;
and we say that the numbers 6, 8, 15, 20 fonn a proportion.
We generally express the fact thus —
6 : 8 : : 15 : 20.
It is easy to find hy trial that the prodtcct of the extreme
terms is equal to the product of the means.
Thus^ we have 6 x 20 = 8 x 15.
We may prove this property of the terms of a proportion
to hold generally as foUows : —
Suppose we have given the proportion 12 : 21 : : 20 : 35.
It follows, from our definition above, that \^ = J J, and
maltiplying each of these fractions by the pi-oduct of their
denominators, viz., by 21 x 35, we have
H X 21 X 35 = 1^ X 35 X 21.
* Now (Art 8), Ji X 21 = V = 12, and }-g x 35 = V - 20,
and we hence have 12 x 35 = 20 x 21.
Now, we have not in our reasoning taken into account the
actual value of the terms of the given proportion ; and it is
thei*efore evident that a similar result will follow from every
proportion, and we may hence conclude generally : —
In every proportion tJie pro(fuct oftliA exiretties U equal iq
62 ARITHMETIC.
39. Having given any tliree terms of a proportion, to find
the remaining one.
Since the product of the extremes is equal to the product
of the means, the following rule is evident : —
Rule. — If the required term be a mean, divide the pro-
duct of the extremes by the other mean ; but if the required
term be an extreme, divide the product of the means by the
other extreme.
Ex. 1. — 28, 24, 30 are respectively the 1st, 3rd, and dth
terms of a proportion, required the 2nd term.
We have — 28 : required term : : 24 : 30
.-. required term = ^%^/° = ^ = 35.
Ex. 2, — 10, 45, 16 are respectively the 1st, 2nd, and 3rd
terms, required the 4th term.
"We have — 10 : 45 : : 16 : required term
.-. required term = ^\^/" = ^-^ = 72.
Ex. 3. — 2 hours, 45 minutes, 8 men are respectively the
1st, 2nd, and 3rd terms, reqidred the 4th term.
We must express (Art. 6) the 1st and 2nd terms in the
same denomination, and the proportion will stand thus —
Min. Min. Men.
120 : 45 : : 8 : required term.
Now, the ratio of the first two terms is the same as the ratio
of the abstract numbers 120 and 45 ; and the 4th term must
be of the same denomination as the 3rd term, otherwise the
3rd and 4th terms could not form a ratio.
We have therefore —
Required term = ^y^ ^®^ = ^f^ ^®^ "= ^ ™^^«
Simple Proportion.
40. In Arithmetic we divide Proportion into Simple and
Compound. Simple Proportion is the equality of two simple
ratios, and therefore contains four simple terms ; and the usual
problem is to find the fourth term, having given the^rs^
three terms.
SIMPLE PROPORTION. 63
When we know the exact order of the given terms, the
fourth term is, of course (Art. 38), found thus —
Rule. — ^Multiply the 2nd and 3rd terms together, and
divide by the 1st
The formal arrangement of the three given terms in their
proper order is called the statement; and the only diiiiculty,
therefore, in working a sum in Simple Proportion, or Single
Hule of Three, as it is called, consists in stating it.
We shall work a few examples to illustrate the mode of
doing this.
Ex. 1. — ^If 12 men earn £18, what will 15 men earn
under the same circumstances ?
We have here two kinds of terms, nien and earnings, and
whatever ratio any given number of men bears to any
second given number of men, it is evident that it must be
equal to the ratio of the earnings of the first lot of men to
the earnings of the second lot, and we may therefore write —
Men. Men.
12 : 15 = £18 : 2nd earnings.
Men. Men.
or 12 : 15 :: £18 : 2nd earnings.
As the first two terms are of the same denomination, their
ratio is not altered by treating them as abstract quantities,
and the denomiimtion of the 4th term must be the same as
that of the 3rd.
Hence we have —
Ans. : = £'-^~^ = £^-V^ = £22. 10s.
Ex. 2. — If 18 men d6 a piece of work in 25 days, in what
time will 20 men do it ?
The two kinds of terms we have here to consider are men
and time. In doing work we know that the tims will dimiiv-
ish exactly as the number o£m>en increases, and hence the ratio
of the second lot of men to the Jirst lot will be equal to the
ratio of the given time to the tirne required. We therefore
have —
Men. Men. Days.
20 : 18 :: 25 : required time.
.-. Ans. : = "Ls.-^. days = «-f^ days = 22 '5 days.
64: ARITHMETIC.
We liave reasoned out the above examples thus to show that
the working of problems in Rule of Three depends upon the
principle of the equality of ratios. Practically^ however,
we proceed as follows : —
Ex. 1. — If 12 men earn £18, what will 15 men earn under
the same circumstances 1
"We are required to find earnings, and we therefore put
down for the 3rd term the given earnings, thus —
£
^' 18
• •
The question is with regard to 15 men instead of 12 men,
and we know their earnings must be greater. We therefore
place the greater of these terms in the 2nd place and the
other in the 1st, and the statement becomes —
Men. Men. £.
12 : 15 : 18 : required earnings.
.•. as before —
Ans. = £^V^ = £^-^ = £22. 10s.
Ex. 2. — If 18 men do a piece of work in 25 days, in what
time will 20 men do it ?
We are required to find time, and we place therefore the
given time, viz., 25 days, in the 3rd place.
Again, the question is with regard to 20 men instead of
18 men. Now, we know that 20 men require less time than
18 men to do a piece of work, and we hence place the less of
these teiTOs in the 2nd place. The statement then becomes —
Men. Men. Days.
20 : 18 : : 25 : required time.
.'. as before —
Ans. : = '-—^ days = ^ days = 22-5 days.
Ex. XV.
1. If 12 articles cost £15, what will 624 costi
2. What is the price of 35 loaves, when 29 loaves cost
15s. 8JdJ
3. If I get 140 metres of cloth for 541 ffr 7^ ^'} what mustj
I pjij^ fpr 8§ wetres, 3 4ecii».'{
COMPOyND PBOPOBTION. 65
4. If 4 cubic. metres of water run into a cistern in 18
minutes, in what time will it be full, supposing it to be 4
metres long, 6 metres, 25 oentim. deep, and 35 decim. wide ?
5. K the carriage of a parcel for the first 50 miles be
Is. 3d., and if the rate be reduced by one-third for distances
beyond, how far can the parcel be carried for Is. 7d.?
6. If a half-kilogram of sugar cost 1 fr. 10 c, what will be
the cost of 3 kilog. 625 grams, t
7. There are two pieces of the same kind of cloth, measur-
ing 43 yards and 57 yards respectively, and the second costs
£1. 9s. 2d. more than the first. What is the cost of the
firsti
8. A garrison of 720 men have provisions for 35 days, and
after 7 days 120 more men arrive. How long will the provi-
sions last)
9. After paying 4d. in the pound income-tax a person has
£299. 18s. 4d. lefk» What was the amount of his original
income 1
10. Two clocks, one of which gains 3 minutes and the
other loses 5 minutes per day, are put right at noon on Mon-
day. What is the time by the second clock when the first
indicates 4 p.m. on the foUbwing Thursday?
11. When will the hands of a clock be exactly 30 minute
divisions apart between 2 and 3 o'clock ?
12. If I lend a friend £120 for 9 months, how long ought
he to lend me £270 1
Compound Proportion.
41, Compound Proportion is an equality between ratios,
one of which at least is a ratio compounded of two or more
simple ratios.
Arithmetical questions depending on Compound Proportion
are generally said to belong to the Double Kule of Three ;
and the proportion consists of an equality between a ratio,
on the one hand, compounded of two or more simple ratios ;
and, on the other hand, a simple ratio, whose consequent 19
required.
The following examples will illustrate the method of ^ork«
jng questions in this r^le :r—
66 ABITHMETIC.
Ex. 1. — ^If 12 horses eat 20 bushels of com in 8 days^ in
what time will 24 horses eat 16 bushels?
24 horses : 12 horses ) « , • j x-
20buahel8: 16 bushels | ' = ^ ^^ '• m^^>^^'^
Explanation. — ^We are required to find tinie ; and so, as
in simple proportion, we put in the 3rd place tiie given
time, viz., 8 days. .
Learning y for the presmty the guantUy ecUen out qfeantidero'
tioriy we know that 24 horses require less time to consume a
given quantity of food than 12 horses do ; we therefore place
the less of these two terms in the 2nd place, and the other
in the 1st place.
(The statement np to this point is 24 horses : 12 horses : s 8 dAjB
: required time, and we might obtain 4 days as an answer, irrespective
of the quantity eaten. We might now place this answer in tae 3rd
term of another simple proportion, and take the quanUty eaten into
consideration, irrespective of the number of horses, thus getting an
answer depending both upon the number of horses and tiie quantUg
eaten. It is more convenient, however, to proceed thus :)
Again, taking into consideration the quantity eateuy and
leaviTig out of consideration the other given pair of terms, we
see that less time is required to eat 16 bushels than to eat 20
bushels, "We, therefore, put the less term in the 2nd place,
and the other in the 1st.
Now, treating the terms of the ratios which occupy the
1st and 2nd places as abstract quantities, and compound-
ing them, we nave :
24 X 20 : 12 X 16 :: 8 days : required time.
.-. required time = ' ^^ i",o " ^J^ = ^'^ ^7^
Ex. 2. — ^How much bread can I get for 9d. when wheat
is at 18s. a bushel, if the fourpenny loaf weigh 3 lbs. when
wheat is at 20s. a bushel %
Proceeding as in Example 1, we have
18 • *>0 I '• 3 lbs. : weight required.
Or, 4 X 18 : 9 X 20 :: 3 lbs. : weight required.
, % weight required ;= ^-^"^ ^>>^. = ^'^ ^^^
COMPOUND PBOPOETION 67
Ex. XVI.
1. If 15 men can build a wall 81 feet long in 18 days, how
many men can build 135 feet of the same kind of wall in 30
daysf
2. In 4 daysy 18 workmen can dig a ditch 162 yards long,
7 feet wide, and 12 feet deep. What must be the depth of a
ditch which 45 workmen can dig in 7 days, supposing it to
be 387 yards long and 5 feet wide 1
3. A traveller,* going 15 hours a day, walks 1500 kilo-
metres in 20 days. How far will he go in 30 days, walking
12 hours a day with the same velocity] Express your
answer in English miles.
4. Two men are partners ; one puts in a capital of £800,
and receives as 6 months' profit £120. What is the capital
of the other, who receives £3375 as 9 months' profit ?
5. Two tourists having spent £1. 16s. 8d. in 2^ days, meet
three others with whom they continue their tour, and they
spend while together £21. Is. 8d., at the same rate per day.
Bequired how long they were in company.
6. If 16 men and 10 boys do a piece of work in 10 days,
in how many days would 8 men and 18 boys do a piece 7
times as great, supposing the work of 5 boys equal that of
2 men)
7. Supposing the rate of carriage to be diminished one-third
after the first 50 nules, find the cost of carrying 16 cwt. for
40 miles, when 12 cwt. can be carried 100 miles for 4s. 2d.
8. A dstem is 8 metres, 4 decim. long, 1 metre, 8 centim.
wide, and 275 centim. deep. Find the depth of another cis-
tern of equal capacity whose length is 7 metres, 2 decim., and
width 11 decim.
9. Persons whose incomes are less than £300 per annum
are taxed upon *£80 less than their income. Supposing 3
persons, having equal incomes, to pay £7 in the aggregate, at
4d. in the pound, find the total tax upon 14 persons, each
having incomes 3 times as great.
10. If 12 horses eat 10 acres of grass in 16 weeks, and 18
horses eat 10 acres in 8 weeks, how many horses would eat
40 acres in 6 weeks, the grass being supposed \^ gcw ^is^-
farmfy-f
68
11. A boat^ propelled bj 8 oan, whidi take 28 skroteB per
minute^ goes at the rate of 9} miles per luyar. Find the rate
of a boat propelled by 6 oars, which take 36 strokes per
minute^ the work done bj each stroke of the latter being
one-sixth less than that by each stroke of the former.
12. If 4 men and 10 women can do a piece of work in 8
day^y which 12 women and 20 children can. do in 4 days, in
what time will 6 men, 18 women and 5 children do a work
three times as great 1
CHAPTER VL
APPLICATION TO ORDINARY QUESTIONS OF 00]CHEBC«
AND TRADE.
Interest.
42> Interest is the money paid for the use of money.
The Principal is the money lent, and the Amount is the
sum of the interest and principaL
The Bate of interest is the money paid for a given sum for
a given time. £100 is in practice the given sum, and one
year the given time. Thus, if £4 be paid for the use of £100
xor one year, the rate of interest is £4 per cent, per annum,
or as we generally say, 4 per cent
Simple Interest is interest calculated on the original
principal only.
Compound Interest is the interest which arises from
adding the interest for each year to the principal of that
year, and calculating interest for the next year upon the
amount so obtained*.
Simple Interests
48. BuLE. — Multiply the principal by the rate per oenl
and by the number of years; then divide the product by
100, and the quotient will be the simple interest,
SIMPLE INTEREST. 69
Ex. 1. — ^Find the simple interest of X420 for 3 years at
5 per cent.
420
5
2100
3
J^Sm Ans.: X63.
Beason for this process —
£100 gains £5 in 1 year, and the question is to find the
gain upon £420 in 3 years.
Hence, proceeding as in Double Rule of Three, we have —
100 X 1 : 420 X 3 :; £5 : interest required.
.'. interest required = £i-io-JULJL_s., which is exactly as
Stated in the rule.
Ex. 2. — ^Find the simple interest of £352. Is. 8d. from
March 16 to August 21, 1873, at 4 per cent.
The foUowing process will be easily imderstood :—
£ 8, d,
352 1 8
4
£U-08 6 8
20
1 GGs.
12
■ 8-OOd.
.'. £14. Is. 8d. is the interest for 1 year.
Now, from March 16 to August 21 are 158 days; hence
we have —
365 days : 158 days :: £14. Is. 8d. : Interest required.
.". interest required = £6. Is. ll/jd.
Ex. XVIL
Find the simple interest of —
1. £350 for 4 years at 5 per cent.
2. £295. 2s. Id. for 3^ years at 4 per cent.
70 ARITHMETIC.
3. £375. 8s. 4(L for 2J years at 4J per cent.
4. £160 from Feb. 1 to June 12, 1872, at 7 J percent,
5. £48 for 7 months at 1 J per cent per month.
6. £219. 4s. 2d. for 6 years at If per cent.
In the six following examples, nndersfcand simple interest
7. At what rate per cent, will £129. 8s. 4d. gain £6. 3s.
5 Jd. in 2^ years ?
8. A certain sum amounts in 3 years at 7J per cent to
£289. 16s. 3^d. ; find the original sum.
9. In what time will £175. 6s. 3d. amount to £192. 7s.
lOJd. at 2 J per cent 1
10. What sum will amount in 2 years 9 months at 4 per
cent to £427. 7s]
11. If £320 gain £9 in 13 months, in what time will
£480 gain £6 at the same rate*?
12. Find the interest of £29. 7s. 5d. for 6 months at 5j\
per cent.
Compound Interest
44. Rule 1. — ^Find the interest for one year as in Simple
Interest, and add it to the principal ; then find the interest
for one year upon this amount reckoned as principal for the
Second year, and add it to the second year's principal^ and so
on. Subtract the original principal from the amount so
obtained for the given number of years, and the result will
be the compound interest required.
Rule. 2. — Divide the given rate per cent by 100, putting
the result in a decimal form, and place wm^i/ before the
decimal point Raise the number thus obtained to a power
corresponding to the given number of years, multiply the
principal by the result, and we get the amount for the given
number of years.
Thus, supposmg 5 to be the rate per cent, we have,
dividing by 100 and placing unity before the result, the
number 1*05. Then, if the given number of years be 4,
and £162 the principal, we have, according to rule, amount
= £162 X (1-05)^
llie first rule requires no explanation; the second rule
may be explained thus : —
Compound interest. 71
Ex. — ^Find the compound interest of X360 for 3 years at
4 per cent.
Now, interest for £100 for 1 year = £4,
„ £1 „ = £-04;
hence, amount of £1 „ = £1 + £'04 = £1*04.
We thus see that the amount of £1 for 1 year at 4 per cent*
is 1'04 times the original sum. It therefore follows that —
Amount of the £1*04 for 1 year = 1*04 times £1-04,
.-. amoimt of £1 • „ 2 years = £(1-04 x 1-04)
= £(1 -04)' ;
and so, amount of £1 „ 3 years = 1*04 times £(1-04)'
= £(l-04)»,
hence, amount of £360 „ „ = 360 times £(1-04)'
= £360 X (1-04)',
The compound interest is then found by subtracting from
this the original principal.
Ex. 1. — Find the compound interest of £570 for 3 years
at 5 per cent.
(We shall work this by Hule 1, and for convenience shall
keep our quantities in a decimal form.)
£
670
5 :
£28-50 = interest foi* first year.
570
£598*5 = principal for second year*
£29-925 = interest „ „
598-5
£628-425 = principal for third year.
5
£31-42125 = interest „ „
628-425
£659-84625 = amount at end of third year,) therefore
570 = original principal ; J subtractings
£89-84625 »■ compound interest for 3 years.
72 XRITHMETia
Ex. 2.— Find the compound interest of £327. 12s, 6d. iot
4 years at 3 per cent.
Now, £327. 12s. 6d. ±: £327-625.
Hence, by Rule 2 —
Amount = £327*625 x (1-03)* = £368. 14s. lOfd. nearly,
Ex. 3. — Wliat sum of money, if put out for 2 years at 4
per cent., will amount to £324. 9s. 7^d., compound interest
being reckoned ?
By Rule 2, the principal may be foimd by dividing the
amoimt by (1 •04)'.
Now, given amount = £324. 9s. 7id. = £324-48.
Hence principal required = £32448 -r (1-04)* = £300.
Ex. XVIII.
Find the compound interest of
1. £284 for 2 years at 4 per cent.
2. £312. 12s. 7^d. for 3 years at 5 per cent.
3. £283. 10s. for 2 years at 3 J per cent
4. £605. 1 2s. 6d. for 4 years at 4 per cent.
5. What is the difference between the simple and com-
pound interest of £150 for 2 years at 6 per centl
6. Find the amount of £381. 1 florin 3 cents 5 mils for
3 years at 5 per cent.
(£1 = 10 florins, 1 florin = 10 cents, 1 cent = 10 mils.)
7. Find the amount of £250 for 2 years at 4 per cent, j^er
annum, interest being payable half-yearly.
8. What sum will amount in 3 years at 4|^ per cent com-
pound interest to £200 ?
9. A town has 200,000 inhabitants, and it increases at the
rate of 5 per cent per annum ; find the number of inhabi-
tants at the end of 3 years.
10. Find the difference in amount of £350 for 3 years at
4 per cent simple interest, and £420 for 2 years at 5 per
cent compomid interest.
1 1. How much would a person who lays by £50 a year at
5 per cent compound interest, draw out at the end of 4 years?
12. A person expects to receive £450 in 3 years ; what
present sum is equivalent to this, reckoning compound inter-
est; at 4 per cent ?
DISCOUOT. 73
Bisconnt.
46. When money is paid before it is due, the payee may,
of course, put out tiie money at interest for the rest of the
term, and thereby increase it. It therefore follows that the
amount which ought to be paid for the discharge of an
account before its proper time should be such a sum that, if
put out at interest for the remainder of the term, will just
amount to the original Rum in question.
Thus JB102. 10s. (interest being reckoned at 5 per cent
per annum) payable 6 months hence, would be fully dis-
charged by paying £100 at once. For £100 in 6 months at
6 per cent, per annum would amoimt to £102. 10s. Hence,
the payee ought to remit £2. 10s. from the full accoimt. The
amount remitted is called discoant.
It will be seen, therefore, that the discount on £102. 10s.
due 6 months hence at 5 per cent is £2. 10s.
Bankers, however, are in the habit of charging interest
instead of discount The banker's discount, therefore, on
£100 due 6 months hence at 5 per cent, is £2. 10s.
Hence, the true discount on £102. 10s. due 6 months at 5
per cent is the same as the hankers discount on £100 under
the same circumstances ; and bankers' discount on any given
sum is in excess of the true discoimt.
Tradesmen's bills are legally due three days after the term
for which they are drawn is completed. This extension of
time is called three dmjs of grace. When a bill falls due on
a Sunday, it is usual in England to meet it on the previous
Saturday.
Ex. 1. — ^Find the difference between the true discount and
the banker's discount on £30 6 due 4 months hence at 6 per cent
Now, £100 would in 4 months gain \ of £6, or £2.
Hence, the true discoimt on £102 due four months hence at
6 per cent is £2, and therefore we have —
£102 : £306 :: £2 : true discount required.
Hence, true discount = £^i^f-* = £6.
Again, proceeding according to the rule for simple
interest —
Banker's discount = j^^ 2 ^ l\\\:=^ £&. \%. K%.
74 ARITHMETia
Hence, tlie excess of the banker^s discount over the true
discount is 2s. 4|d.
Ex. 2. — A bill of £350 drawn on March 15, at 6 months, is
cashed on May 20, 1872 ; what is the banker's discount at 6
per cent. ?
The bill is legally due on Sept 18, and from May 20 to
Sept. 18 are 121 days.
Now, the interest on £350 for i. year at 6 per cent, is easily
found to be £21. Hence,
365 days : 121 days :: £21 . banker's discount required ;
and /. banker^s discount = £^^~^ = £6. 19s. 2j^5d.
Percentages.
46. There are many questions which relate to ordinary
coromercial transactions which may be worked exactly as
if we had to find the simple interest for one year — e.g,, ques-
tions in commission, brokerage, insurance, &c»
Commission is a sum of money charged, by an agent for
buying or selling goods, at a certain rate per cent, upon the
value of the goods.
Brokerage is simild,r to commission^ but it is charged upon
money transactions instead of upon the sale of goods.
Insurance is a sum charged per cent upon the value of
property, the said value being paid to the insured in case of
loss from causes as per agreement.
Ex* 1.— ^Find thd brokerage on £625; 5s. at 4J per cent.
£ 8, d*
625 5
H
^1^.^ I .li^o^^^ 111 Tl
2501
312 12 6
28-13 12 6
20
2 -728.
12
8-70 Ana. \ BfL%. 1^ %^^d.
PERCENTAGES. 75
Ex. 2.— The rate of insurance is 4 J per cent., and the
"value of some property insured is worth £766. What will
be the annual payment, so that, in case of fire, the owner
may receive back his premium, as well as the value of his
property ?
K, instead of paying £4. 5s. as insurance on every £100,
he pays £4. 5s. upon every (£100 - £4. 5s.) or upon every
£95. 15s. ; then, in case of fire, he will receive £100 for a
damage of £95. 15s., and thus have the value of his property
and the amount of his premium.
The problem is, therefore —
If £4. 5s. is the {Premium on £95. 15s., what is the pre*
mium on £766 1
Hence —
£95. 15s. : £766 : t £4. 5s. t premium required.
And, therefore^ premium = £34*
1. VmA the hankered discoilnt On £412^ dud 6 months)
hence, at 6 pe^ cent»
2. By how itiuch does the bankdi^'s discoimt on £100, due
3 montjis hence, at 5 per Centi^ exceed the true ? _
3. What discount would bd charged upon a bill drawn for
£320, on April 15, at 4 months^ and presented for payment
on June 3 (discount at 7 per cent.) ?
4. Find the discoimt on a bill drawn oil Aug. 3 for £200j
at 6 months, and cashed on Sept. 10, discount being reckoned
at 6 per cent^
5. A man buys goods for £250, being allowed 6 months
credit, and he immediately sells them for the same amoimt,
allowing 3 months credit. What does he gain by the trans-
action, interest being reckoned at 5 per cent. ?
6. Find the brokerage on £352. 17s. 6d. at 3 per cent.
7. What is the brokerage on £4500 at ^ per cent. ]
8. What is the commission on the sale of goods to the
amount of £850 at 5 per cent.?
9. A person insures for £1050 at 3| per cent. "WaaX* \^
}u8 annual premium f
76 JLBITHUETIC
10. What will be the annual premitun t)n property worth
£965, 80 that the insured may obtain his premium bac^
again with the value of his property, in case of loss — ^ini^^rance
being at 3^ per cent. ?
11. The brokerage on ascertain sum at ^ per cent, is
£5. 10s. 7^d. Find the sum.
12. Together with a commission of 4 per cent., goods cost
a person £339. 8s. 5d. Find the cost price to the agent.
Stocks and Shares.
47. When a large amount of capital is to be raised, a
company is generally formed, which raises the money by the
issue of shares. We will suppose a person to hold a £100
share ; he will then be entitled to such a part of the profits
of the company as £100 is of the whole capital. If there bo
a great demand for these shares, persons holding them may
dispose of them for more than the nominal value, say for
£106 ; whereas, if they are very little in demand, the seller
may be glad to sell at, perhaps, £70. In the first instance,
we should say that the shares were at 106, or at 6 premium;
and in the second instance, that the shares were at 70, or at
30 discotmt. If the selling price of the shares is £100, they
are said to be at par.
So, when we read that the Three per Cent Consols are
quoted at 96|, it means that an acknowledgment of indebted-
ness on the part of the Government to the amount of £100,
bearing interest at 3 per cent, per annum, may be bought for
£96|. . ^
The buying and selling of stocks and shares is carried on
by brokers, who charge a percentage from J to J, sometimes
upon the nominal value of the stock, but mostly upon the
actual cash value. When brokerage is to be taken into ac-
count in any example it will be mentioned, and it will
be estimated by the first method, unless specified.
Ex. 1.— What is the value of £1000 stock at SQjV per
cent?
No. of cents, stock = ^-^ = 10
value required = £89^^ x\Q = X'^^^-^*.^^
# •
STOCKS AND SHARES. 77
Ex. S.^What would be the cost of £1180 stock at 153^
per cent, including brokerage at ^ per cent 1
No. of cents, stock = \'^ = ^,
and total cost of each cent. = £(153^ + i) = £153 J.
Hence, required cost = £153j x -^-^-^ = £1811. 6s.
£x. 3. — What is the annual income arising from investing
£6510 in the Four per Cents, at 93 1
Here, price of £100 stock = £93,
aiid .-. No. of centa stock for £6510 = ^f^ = 70.
Henoe^ annual income = £4 x 70 = £280.
Ex. XX.
1. Find the cost of £750 stock at 92| per cent
2. I sell out £325 Three per Cent Consols at 94. What
do I get after allowing the broker \ per cent upon the cash
lie receives for the sale 1
3. Invest £5065. 9s.. 9d. in the Three per Cent Consols
at91f.
4. What would be the cost of £413. Is. 9d. reduced Three
per Cents, at 92}, including a brokerage of \ per cent upon
the cost to the broker ]
5. What would be the proceeds of the sale of £6228 India
Five per Cent stock at 111^, deducting ^ per cent, upon the
sellmg price for brokerage ?
6. If I sell £8160 Spanish Three per Cent Bonds at 31
per cent, and invest the proceeds, less ^ per cent, brokerage,
in Indian Eailway Five per Cent Stock at 107, what will
he the difference in my annual income 1
7. What is the value sterling of $4000 American Bonds
at 93f per cent ($ = 4s. 6d.)1
8. What is the value of $37,000 United Btatea lBoi^^\x\»
78 ABITHMETIC.
9. Invest £954. Os. 7d. in India Eive per Cent Stock at
110}, allowiDg ^ per cent, brokerage.
10. What is the sterling value of 5000 francs Italian
Bonds at 67 per cent. (Exchange 25 fr.)?
11. Find the sterling value of 6000 guilders Dutch 2\ per
Cent Bonds at 56:^ (Exchange 12 guilders).
12. Invest £1025 in French Rentes at 51 J, allowing J per
cent brokerage (Exchange 25 fr.), ^
Annuities,
48, Annuities are annual payments, the first payment
being due at the end of a year. When an annuity is left
untouched for a number of years, its amourU is properly ob-
tained by allowing compound interest,
49. To find the amount of an annuity of a given sum for
a given time, at a given rate per cent
Let us suppose the annuity to be £1, the time 4 years,
and 5 the rate per cent. The first payment will not be due
till the end of a year ; so that at the end of 4 years it will
have been accumulating for 3 years at compound interest;
so the next payment, not being due till the end of the second
year, will have been accumulating for 2 years at compound
interest ; and so on, the last payment being made when due.
Hence, the amount of an annuity of £1, for 4 years at 6
per cent, will be (reversing the above order) as follows : —
£1 + amount of £1 for 1 year + amoimt of £1 for 2
years + amount of £1 for 3 years (compound interest being
reckoned).
And it is evident that for any other annuity we may follow
the same method, and at the end multiply the sxun by the
number of £ in iJie annuity.
Ex. — Find the amount of an annuity of £300 for 4 years
at 5 per cent.
We aball find the respeoftivQ omwvmts by Rule 2, Art 44.
Thus—
ANNUITIES. 79
£1*05 = amount for 1 year of £1 at 5 per cent.
1-05
525
105
XI -1025 = „ „ 2 years „ „
1-05
55155
11025
M57625 = „ „ 3 years „ „
Hence, amount of annuity of £1 for 4 years —
= £1 + £1-05 + £1-1025 + £1-157625 = £4-310125.
.-. amount of given annuity = £4-310125 x 300 = £1293.
Os. 9(1
60. To find the present value of an annuity to continue
for a given time.
By the present value of an annuity to continue a given
nmnber of years is meant such a lump sum which, paid down
at once, would, by accumulating at compoimd interest for the
same time, amount to just the same sum as the annuity itself
if it were allowed to accumulate. In the absence of alge-
braicid symbols, we shall best illustrate the method of finding '
such a lump sum by an example.
Ex. — ^Find the present value of an annuity of £50 for 3
years at 4 per cent.
Ab in Art 49, we find the amount of the annuity —
= {£1 + £1-04 + £{l'Oiy\ X 50 = £156-08.
Now, whatever be the present value required, we know
that its amount in 3 years at 4 per cent, compound
interest is found (Art, 44) by multiplying it by (1-04)* or
M24864.
It, therefore, follows that if we know this amount before-
hand we can find the present value by dividing it by
1124864.
Hence, present value = £156-08 ^ 1-124864 = £138-755
nearly.
51. To find the present value of 9in. annuity \^ qotl\)\£^<^
for even
80 ARITHMETIC.
It is evident that the sum we require is one which, put
out at interest, will annually produce a sum equal to that of
the annuity itself. The problem then is simply this —
Having given the interest for 1 year of a certain sum,
and the rate per cent., to find the principal
"We have therefore the following rule : —
Rule. — Divide the given annuity by the rate per ceuu:,
and multiply by 100, and the result is ^e present value.
Ex. — How much must a gentleman invest at 5 per cent, in
order to endow a charity with £60 a year.
Present value of the annuity of X60 to continue for ever —
= £^^^ = £1200.
There are many other questions connected with annuities
which are, however, best left till the student has a knowledge
of Logarithms.
Ex. XXL
Find the amount of an annuity of —
1. £120 for 3 years at 4 per cent.
2. £250 for 4 years at 4j^ per cent.
3. £321 for 5 years at 6 per cent.
4. What is the present value of an annuity of £80, to
continue for six years, at 6 per cent. 1
5. A person who, according to the tables of mortality, is
likely to live 10 years, wishes to insure an annual payment
of £40 during life. What sum must he pay down, reckon-
ing interest at 5 per cent. 1 (Give the result to four places
of decimals.)
6. A house produces a clear rental of £30. How many
years' purchase is it worth, interest being reckoned at 5 per
cent.1
7. A gentleman invested a sum of money in the Three
per Cent. Consols, in order that an annual payment of 7s. 6d.
a year might be made in bread for ever. What sum did he
invest)
8. Find the present value of a pension of £120 a-yeai:,
pat/ahh Judf -yearly for 5 years, interest being at the rat^
of 5 per cent per ajanum.
PROFIT AND LOSS. 81
9. A house, which ordinarily lets for £80 a-year, is leased
for a term of four years, at a rent of £20, a certain sum
being paid in addition at the time of letting. Find this
Jatter amount.
10. What is the present value of a freehold which pro-
duces a clear rental of £50, but which cannot be entered upon
fin* two years, reckoning interest at 5 per cent.)
11. Knd "die annuity which in four years, at 4 per cent.,
will amount to £100.
12. A corporation borrows a sum of £3000 at 4 per cent.
What annual payment will clear off the debt in ten years )
(Give the result correct to four places of decimals.)
Profit and Loss.
52. All questions involving the loss or gain per cent, by
any transaction belong to this rule, and may be generally
worked by Proportion.
Ex. 1. — ^A man buys goods at 5s. and sells them at 5s. 8d.
Find his gain per cent.
The actual gain upon 5s. is 8d., and we are required to
find the gain upon £100.
Now 5s. : £100 : : 8d. : gain upon £100,
... gain upon £100 = £^f^ = £13i,
or, required gain per cent = ld|.
Ex. 2. — ^By selling goods at 6s. 3d. there is a gain of 25
per cent What will be the selling price to gain 10 per
oenif
Now, selling price of goods which cost £100, so as to gain
25 per cent, is £125, and that to gain 10 per cent, is £110.
Hence £125 : £110 : : 6s. 3d. : selling price required ;
from which, selling price required = 5s. 6d.
Ex. 3. — ^Find the cost price when articles sold at Is. 9d.
mtail a loss of 12^ per cent.
Now, articles which cost £100 when sold at a loss of 12 J-
per cent, must sell for £87^.
Henoe £87^^ : Is. 9d. : : £100 : cost price required •,
i5vi» which, eoat price = 2s,
82 ABITHMETIC.
Ex. XXII.
1. Find the cost price of goods which are sold at a loss of
10 per cent, for 4s. 10^. *
2. Croods which are sold for 7s. lid. entail a loss of 5 per
cent. What should be the price to gain 30 per cent.1
3. A tradesman reduces his goods 7^ per cent. What
was the original price of an article which now fetches
£1. 7s. 9d.1
4. In what proportion must tea at 4s. 2d. be mixed in&,
tea at 6s. a poimd, so that a grocer may sell the mixture at
5s. 6d. and gain by the sale 10 per cent.?
5. A quantity of silk, after paying a duty of 12^ par cent,
cost £54:, Find the original cost price.
6. An innkeeper buys 37^ gallons of brandy at 148. a
gallon, and adds to it sufficient water to enable him to sell it
at the same price and gain 12 per cent. How much water
does he add ?
7. By selling goods at 8s. 2d. a tradesman gains 16f per
cent. What will be the gain or loss per cent, by selling at
6s. Ijd.
8. A company has a capital of £750,000, and the working
expenses for the year have been £42,123. 12s. 6d. What
must have been the gross receipts in order that the share^
holders may receive a dividend of 4 per cent.?
9. If stock which is bought at 91^ is immediately sold at
91 f, what is the gain per cent.?
10. A person buys goods at 6 months' credit and sells
them for cash at the nominal cost price immediately. What
is his gain per cent. ? (Interest 5 per cent.)
1 1. Goods are marked at a ready-money price and a credit
price allowing 12 months. The credit price is X4, 9s. 3d.,
what is the ready-money price ?
12. Goods are now being sold at 10 per cent. loss, i How
much per cent, must be put upon the selling price in order
that they may be sold at 20 per cent, gain ?
Sqoare Boot and Cube Boot.
S3. To avoid unneoeaaary repetition, the student is referred
t« the articles on lavolntion, Algebru, stage I., irhere tbo
iritbmetical principles and methods are expUined.
Dstiinates.
64. The following Bpecimeng %rill give the student an idea
t wliat he may expect to meet with under the head of Esti-
nates. It ia usual, in ordinary ttHnsadaona, to use certain
Itbreviations j as cub. for cubio measure, aup. for superficial
MBBure, run. for running or lineal measure. Builders, too,
■n in the habit of calling twelfths of a. foot — whether it bo
nincal, superficial, or lineal measure — hj the name of inches.
Dm names Tarda, feet, inches, are often written thus;
Ex. 1. — Bigger, Bbicklayer, and Mason.
pol
trUnmor urchoa Ut luftrtbs, , .
' •■■ - ■■— - on Edra, pivtng 1
quErrrw i
DDpEUf bricks In c
Eiin onir lo ipltf ed bdei inglt
uid ifrindawBp
Do. ta akllbg CO
d eavu ol b«l rad bilcki.
1" X 10", «.
1" X 1!"
ARITHHETIC,
Ex. 2. — Cabfekteb and Joiner. (A square = 100 Bq.ft.}
'Sq. Ft. 1 ~
lontroIiiB lo Irfpunor hiirths, ,
bgidi In 11 brink -
- . lo 4 do. V (T wl
Lubaiif in planliw root utd floontlmben,
Do. In Btap ohiuniiirlag eigm ol do.,
7" X 11" nige boutl,. , . ,
1" X ir chmnlsred Ollet to uvea
Inoh fliiui red dukl buHen tor boudcd lloo
r" ie l" locus Hlfirtjoff, plnffiEBd,.
I Ubrnir li) mitred nuigini to bearUiB, . ,
LflmiltflstoBklrtlnff, i
[neb deal trovla ind rivn, jAoufbed^
ton^od, and icrewod on 3 — 7" ^ "*'
carriaffiH, -,,..-. -
L^" waU-Btrin; houud, for tnadi
3^ X H' rounded oik buul-iii], Fimoh-
poUBhed,.., ..,..-..........,-..,.
B" X 1" beaded la«i«
^ — panel doore, bead, fliuh, and ■
4" X 3° rebated uj boded Inune,'.'
41- X T do. do.,
3^ X 11" moulding, mitred,
deal-caied fmnos, wt^ oak ninl
weaClierei] aiUe,
J* mitred bead, .'.
61" X i" bead lining
11 Inoh-beaded csntre boards,
Totil,
MISCELLAKSOUS EXAMPLES. 85
MI8CELUNE0US EXAMPLES.
{^dededfrfmi Univemty and other FxamiruUion Papers.)
1. Show by an easy example that the division of one
wliole nnmber by another is equivalent to a series of
Babtractions.
Divide 102 hy ^ o£ 144.
2. If the Three per Cents, are at 91 ^^ what interest does
this give on £1 00 1 (Omit brokerage and fractions of a penny. )
3. How many lbs. in '321875 of a ton weight? Convert
it into kilograms ^omitting fractions), assuming that a cubic
decimetre of distilled water weighs 15432*35 grains.
4. Beduoe to their simplest forms — -, — — 7=.— and
. J3'b - V2I
sf5'l2 + 4^03375
^80- ^fm
5. Convert i^^ into a decimal fraction, and find the vulgar
£raction corresponding to the recurring decimal '22297.
6. l%ow, by proper attention to the value of the figures, in
IDiiltiplying one number by another, that the order in which
tlie figures of the multiplier are taken is of no importance.
Multiply 61-143 by 47'982 correctly to three places of
decimals, beginning with the left hand figure of the multiplier,
and use as few figures as possible.
7. Extract the square root of 1095*61, and find to three
places of decimals the value of — ;
V^ - r
8. Find the compound interest of £55 for one year, pay-
able quarterly, at 5 per cent, per annum.
A person bought into the Three per Cents, at 98, and
alter receiving tiiree years' interest he sold at 90. How
much per cent, on the sum invested did he gain or lose ?
86 ARITHMETIC.
9. Three gardeners working all day can plant a field in 10
days ; but one of them having other employment can only
work half time. How long will it take them to complete
the work ?
10. What fraction of a crown is | of 6s. 8d. ? What is
the value of f of a guinea ? Reduce llf d. to a decimal of a
pound, correct to five places of decimals.
11. Reduce the expressions —
^5 + T*T + tV - h and V(|^ "^ f^}'
Multiply 49|f by 5(Vff, and add ^^-g to the result
Divide (2^)» - 1 by (2^)« + 3,^^. .
12. A bankrupt's estate amounts to £910. 3s. 1^., and his
debts to £1876. What can he pay in the pound? and what
will a creditor lose on a debt of £57 ?
13. A person having invested a sum of money in the Three
per Cent. Consols received annually therefrom £233, after
deducting the income-tax of 7d. in the poimd. What is the
sum of money] What can the stock be sold for when Consols
are at 94|^.
14. Find the value of :i>oj_^ioo^, and of iigjffy.
15. Prove the rule for finding the value of a circulating
decimal, and divide 4*367 by the circulating decimal -052.
Reduce to its simplest form the quantity ^
8 - ^20
""8 + ^20
16. Three persons, A, E, C, hold a pasture in common,
for which they are to pay £30 per annum* A put in 7
oxen for 3 months ; B, 9 oxen for 5 months ; and C, 4 oxen
for 12 months. How touch rent ought each to pay?
17. Calculate to four places of decimals the value of the
, 4 of '31416
GjqyreBsion 7^=—
MISCELLANEOUS EXAMPLES. 87
18. Fiiui the least common multiple of 16, 24, and 30, and
explain the method.
19. What should be the price of English standard silver,
3740ths fine, in order that the par of exchange between
Hogland and France should be 25 fr. 22 c. — 200 feancs being
corned fix)m 1 kilogram of silver, 9-lOths fine? (1 kilog.
= 15-434 grains).
20. A person buys 100 shares in a company for £3,500 ;
after receiving four half-yearly dividends of 15s. 4d., 20s.
lOd., 30s. 4d., and 38s. 9d. per share, he sells at a profit of
43 per cent. ; reckoning the simple interest of money at 4 per
cent, how much above that interest has he gained?
21. The price of Three per Cent. Consols is 90| ; what sum
must be invested in order to purchase £24 per annum ; and
what is the rate of interest on the money invested ?
Three partners in trade contribute respectively the
sums of £438, £292, £730, with the agreement that each
^as to receive 5 per cent, on their respective investments,
and that the remainder of the gains of the firm, if any, was
to be divided between them in the proportion of the sums
originally advanced. The whole gain of the firm was £200.
What was each man's share ?
23. If 25 tons of goods are purchased for £37. lOs. and
X)ld at 35s. a ton, what is the gain per ton %
At what rate per ton should the goods have been sold
n order to obtain a profit of £9. 7s. 6d. ?
24. Find the value of t*t of £3. 12s. lljd. ; and find the
taction that 3 miles, 2 fur. 100 yards is of 12 leagues, 2
or. 20 yards.
25. The sum of £9040. 16s. is placed in the Three and a
lalf per Cents, at 94 ; find the income obtained, allowing
a the stock purchased ^th per cent, to the broker, and ^J^
er cent, for other expenses.
26. Express as a fraction '200123, and ex]^teB& ^ «i
xmrriog decimal '012 -t '00132,
88 ARITHMETIC.
27. By the reduction of the income-tax from 7d. in the
pound to 5d. a person saves £28. 2s. 6d. a year ; Tfhat is his
income?
28. If 81 bushels of wheat are consumed by 56 men in 5
dayS; how long will 16 men take to consume 28 bushels 1
29. Find the square root of -1, and prove that v '694 = 83.
30. The periods of three planets which move uniformly in
circular orbits round the sun are respectively 200, 250, and
300 days. Supposing that their positions relative to each
other and to the sun to be given at any moment, determine
how many days must elapse before they again have exactly
the same relative positions.
• V •
SECTION II.
GEOMETRY.
EUCLID'S ELEMENTS, BOOK I.
Definitions.
1. A point is that which has position, but not magnitude.
2. A line is length without breadth.
3. The extremities of a line are points.
^- A straight line is that which lies evenly between its
extreme points.
^- A superficies (or surface) is that which has only length
and breadth.
6- The extremities of a superficies are lines.
7. A plane superficies is that in which any two points
being taken, the straight line between them lies wholly in
^t superficies.
S. A plane angle is the inclination of two lines to one
wiother in a plane, which meet together, but are not in the
same direction.
9. A plane rectilineal angle is the inclination
of two straight lines to one another, which meet
together, but are not in the same straight line.
Note. — ^When several angles are at one point B, any one of them is
ezpressed by three letters, of which the middle letter is B, and the
/irst letter is on one of the straight lines which coutava WiQ «i^<^)
and Hhe last letter on the other line.
DO
OEOMETBT.
Thus, the angle contained by the straight lines AB and BC is ex-
pressed either by ABC or CBA, and the angle contained by AB and
BD is expressed either by ABD or DBA. When there is only one
an^le at any given point, it may be expressed^by the letter at thit
point, as the angle E.
1 0. When a straight line standing on another
straight line makes the adjacent angles equal
to one another, each of the angles is called a
right angle; and the straight line which stands
on the other is called a perpendicular to it
11. An obtuse angle is that which is greater
than a right angle.
12. An acute angle is that which is less
than a right angle.
13. A term or boundary is the extremity of anything.
14. A figure is that which is enclosed by one or more
boimdaries.
15. A circle is a plane £gure contained by
one line, which is (»Iled the circU2nfer8nC6^
and is such, that all straight lines drawn
from a certain point within the figure to the
circumference are equal to one another.
16. And this point is called the centre of the circle, [and
any straight line drawn from the centre to the circumferenoe
is called a radius of the circle].
DEFINITIONS. 91
17. A diameter of a circle is a straight line drawn thi*ough
I the centre, and terminated both ways by the circumference.
18. A semicircle is the figure contained by a diameter and
the part of the circumference cut off by the diameter.
19. A segment of a circle is the figure contained by a
straight line and the part of the circumference which it
cuts off.
20. Rectilineal ifignres are those which are contained by
straight lines.
21. Trilateral flgnres, or triangles, by three straight lines.
22. Quadrilateral figures, by four straight lines.
23. Multilateral figures, or polygons, by more than four
straight lines.
24. Ofthree-sided figures an equilateral triangle /
is that which has three equal sides. /__
25. 'An isosceles triangle is that which has only
two sides equal.
/ V
26. A scalene triangle is that which has three
unequal sides.
27. A right-angled triangle is that which has
a light angle.
28. An obtuse-angled triangle is that which
las an obtuse angle.
29. An acute-angled triangle is that which
has three acute angles.
92 GEOMETKT.
30. Of four-sided figures, a square is that w
has all its sides equal, and all its angles right auj
31. An oblong is that which has all its ai:
right angles, but not all its sides equaL
32. A rhombus is that which has all its s
equal, but its angles are not right angles.
33. A rhomboid is that which has
opposite sides equal to one another, but aU
sides are not equal, nor its angles right angl
' 34. Parallel straight lines are such as
in the same plane, and which being prodi
ever so far both ways do not meet.
35. A parallelogram is a four-sided figure of which
opposite sides are parallel ; and the diagonal is the strai
line joining two of its opposite angles. All other four-si
figures are called trapeziums.
Postulates.
1. Let it be granted that a straight line may be drawn fi
any one point to any other point.
2. That a terminated straight line may be produced to ;
length in a straight line.
3. And that a circle may be described from any centre
any distance from that centre.
Axioms.
1. Things which are equal to the same thing are equal
one /mother.
2. If equals be added to ec\v\»\s ^a\lG ^VcJvsa ^x^ ^w^oaL
i
I
EXPLANATION OF TERMS AND ABBREVIATIONS. 93
3. If equals be taken from equals the remainders are equal.
4. If equals be added to unequals the wholes are unequal.
5. If equals be taken from unequals the remainders are
unequal.
6. Things which are double of the same are equal to one
another.
7. Things which are halves of the same are equal to one
another.
8. Magnitudes which coincide with one another, that is,
vbich exactly fill the same space, are equal to one another.
9. The whole is greater than its part.
10. Two straight lines cannot inclose a space.
11. All right angles are equal to one another.
12. If a straight line meet two straight lines, so as to
make the two interior angles on the same side of it taken
together less than two right angles, these straight lines being
oontmually produced shall at length meet on that side
on which are the angles which are less than two right
angles.
Explanation of Terms and Abbreviations.
An Axiom is a truth admitted without demonstration.
A Theorem is a truth which is capable of being de-
monstrated from previously demonstrated or admitted
traths.
A PoBtulate states a geometrical process, the power of
effecting which is required to be admitted.
A Problem proposes to effect something by means of
admitted processes, or by means of processes or construe^
tionSy the power of effecting which has been previously
demonstrated.
A Corollary to a proposition is an inference which may be
easily deduced &om that proposition.
The sign = is used to express eqriality,
4 mmm angle, md A signifies triowjfle.
94
GEOMETRY.
The sign >- signifies " is greater than," and
than."
"is less
+ expresses addition ; thus AB + BC is the line
whose length is the 8um of the lengths of
AB and BC.
— expresses subtraction; thus AB - BC is
the excess of the length of the line AB^bove
that of BC.
AB^ means the square described upon the
straight line AB.
From cen-
tres A and
B, and ra-
dius =AB,
describe
circles.
AC=AB.
BC=AB.
ACandBC
each=AB.
AC=BC.
/.AB=BC
=CA.
Fropositioxi 1.— Problem.
To describe cm equilateral triangle an a given finite straight
line.
Let AB be the given straight line.
It is required to describe an equilateral triangle on AB.
CoNSTRUCJTiON. — From thecentre
A, at the distance AB, describe the
circle BCD (Post.* 3).
From the centre B, at the dis-
tance BA, describe the circle ACE
(Post. 3).
From the point C, in which the
circles cut one another, draw the
straight lines CA, CB to the points A and B (Post. 1).
Then ABC shaU he cm equilateral triangle.
Proof. — Because the point A is the centre of the circle
BCD, AC is equal to AB (Def. 15).
Because the point B is the centre of the circle ACE, BC
is equal to BA (Def. 15).
Therefore AC and BC are each of them equal to AB.
But things which are equal to the same thing are equal to
one another. Therefore AC is equal to BC (Ax. 1).
Therefore AB, BC, and CA are equal to one another.
Therefore the triangle ABC is equilateral, and it is de-
scribed on the given straagiit \me i^« Wlxwifi. -was to be done.
PB0P08ITI0NS. 95
Proposition 2.— ProUein.
From a given point to draw a straight line equal to a given
ttndght line.
Let A be the given point, and BC the given straight line.
It ia required to draw from the point A a straight line
e^toBC.
GoNSTBUcnoK. — ^From the point A to B draw the straight Draw ab.
line AB (Post 1).
Upon AB describe the equilateral triangle DAB (Book I., A dab e-
Prop. 1). quiUteraL
Produce the straight lines DA, DB, to E and F (Post. 2).
.. From the centre B, at the dis- ^ — ^ Se"**"**
I tance BC, describe the circle CGH, ^'
meeting DF in G (Post. 3).
From the centre D, at the dis- I /^^'iT^X \ Dwceu-
tance DG, describe the circle GKL, « ^ i^ a\ I tw.
meeting DE in L (Post. 3).
Then AL shall be eqvxd to BC.
Proof. — Because the point B is ^^^^^J^^^^ bc=ijg.
tiie centre of the circle CGH, BC is
equaltoBG(Def. 15).
Because the point D is the centre of the circle GKL, DL dl=dg.
is equal to DG (Def. 15).
But DA, DB, parts of them, are equal (Construction). da=db.
Therefore the remainder AL is equal to the remainder BG -^^ =
(Ax. 3).
But it has been shown that BC is equal to BG.
Therefore AL and BC are each of them equal to BG. iib'^^'.?"^^
But things which are equal to the same thing are equal to bo.*^*^ ~~
one another, therefore AL is equal to BC (Ax. 1).
Therefore from the given point A a straight line AL has .*. al =■
teen drawn equal to the given straight line BC. Whicli was ^^'
^ he dime.
Proposition 3.— Problem.
From the greaier of tioo given straight lines to cut off' a part
^jual to tfie less.
Let AB and C be the two given straight lines, of which
Afi is the greater.
96
OEOMETBY.
Make AD
A ai centre
and radius
AD.
AE=AD.
AD=C.
AEandC
each=AD*
.-, AE=C.
It is required to cut off from AB, tlie greater, a part &
to C, the less.
Construction. — ^From the poii
draw the straight line AD equal i
(I. 2).
From the centre A, at the disti
AD, describe the circle DEF, cut
AB in E (Post. 3).
Then AE shall be eqvxd to C.
Proof. — Because the point A is the centre of the c:
DEF, AE is equal to AD (Def. 15).
But C is also equal to AD (Construction).
Therefore AE and C are each of them equal to AD.
Therefore AE is equal to C (Ax. 1).
Therefore, from AB, the greater of two given straight \\
a part AE has been cut o% equal to C, the less. Q, E, .
AB=DE.
AC=DF.
ZBACa
^EDF.
Proposition 4.— Theorem.
If two triangles Iiave two sides of the one equal to two <
of the other y ea/ch to each, and have also the a/ngles conta'
by those sides equal to one another : they shall have their h.
or third' sides, equal; a/nd the two tria/ngles shall be equal,
their other angles shall be equal, each to each, viz,, thoa
which the equal sides are opposite. Or,
If two sides a/nd the contained angle of one triangle be
spectively equal to those of another, the triangles are eqwk
every respect
Let ABC, DEF be two triangles which have
The two sides AB, AC, equal to the two sides DE, !
each to each, viz., AB equs
A P DE, and AC equal to DF.
And the angle B AC equi
the angle EDF : — ^then —
The base BC shall be e(
to the base EF ;
The triangle ABC shall
equal to the triangle DEF
* Q. E, F. is an abbremtionloT (jiwcC era* /acie?M?tw», that is " «
fpos fg be done,^*
PROPOSinoKs. 97
And the other angles to which the equal sides are opposite,
shall be equal, each to each, viz., the angle ABC to the angle
DBF, and the angle ACS to the angle DFE.
Proof. — ^For if the triangle ABC be applied to {or placed Suppoie
tt/xm) the triangle DBF, ^ abo
So that the point A may be on the point D, and the ST dS?"
stiraight line AB on the straight line D£,
The point B shall coincide with the point E, because AB
is equal to D£ (Hypothesis).
And AB coinciding with DE, AC shall coincide with DF,
because the angle BAC is equal to the angle EDF (Hyp.).
Therefore also the point C shall coincide with the point F,
because the straight line AC is equal to DF (Hyp.).
But the point B was proved to coincide with the point E.
Therefore the base BC shall coincide with the base EF.
Because the point B coinciding with E, and C with F, if
the base BC do not coincide with the base EF, two straight
lines would enclose a space, which is impossible (Ax. 10).
Therefore the base BC coincides with the base EF, and is bo=ef.
therefore equal to it (Ax. 8).
Therefore the whole triangle ABC coincides with {he whole .*. A ab<
triangle DEF, and is equal to it (Ax. 8). = A dei
And the other angles of the one coincide with the remain- z abc =
ing angles of the other, and are equal to them, viz., the angle ^ aot*=
ABC to DEF, and the angle ACB to DFE. i DFE.
Therefore, if two triangles have, &c. (see Enunciation).
Which was to he shown.
Proposition 5.— Theorem.
The cmglea cat the base of a/n isosceles tria/ngle are equal to
one another ; and if the eqital sides be prodv^ced, the angles upon
the other side of the base shaU also be eqtuil.
Let ABC be an isosceles triangle, of which the side AB ia ab=ac,
equal to the side AC.
Let the straight lines AB, AC {the equal sides of the tri-
angle)y be produced to D and E.
The angle ABC shall be equal to the angle ACB (^vagjUa
aitheda^e),
S Q
98 GEOXETBT.
And the angle CBB shall be equal to the angle BCE
{a/rufles upon the other side of the hose).
CoNSTBUcnoN. — In BD take any point
F.
AG = AP. / \ From AE, the greater, cut off AG,
equal to AF, the less (I. 3).
Join FC, GB.
Proof. — ^Because AF is equal to AG
i Construction), and AB is equal to AC
Hyp.),
FA, AC re- / \ Therefore the two sides FA, AC are
^a7ab. i/ ^, equal to the two sides GA, AB, each to
each;
And they contain the angle FAG, common to the two
triangles AFC, AGB.
/.FC=GB Therefore the base FC is equal to the base GB (I. 4);
= A AOT? -^^ *^® triangle AFC to the triangle AGB (L 4) ;
And the remaining angles of the one are equal to the
z ^G.~ remaining angles of the other, each to each, to which the
z AFC = equal sides are opposite, viz., the angle ACF to the angle
And because the whole AF is equal to the whole AG, of
which the parts AB, AC, are equal (Hyp.),
BP=CG. The remainder BF is equal to the remainder CG (Ax. 3),
And FC was proved to be equal to GB ;
Therefore the two sides BF, FC are equal to the two sides
CG, GB, each to each.
And the angle BFC was proved equal to the angle
CGBj
Therefore the triangles BFC, CGB are equal; and their
other angles are equal, each to each, to which the equal sides
are opposite (I. 4).
/. z FBO Therefore the angle FBC is equal to the angle GCB, and
zBCP =' the angle BCF to the angle CBG.
z CBG. ^(j since it has been demonstrated that the whole angle
ABG is equal to the whole angle ACF, and that the parts of
these, the angles CBG, BCF, are also equal,
- ^ACB Therefore the remaining angle ABC is equal to the remain-
'" ' ing angle ACB (Ax. 3),
Which are the angles at ^^ \)^^ oi ^<b Xjcvsnc^^'^ KS»Ci^
PROPOSITIONS. 99
And it has been proved that the angle FBC is equal
to the anglB GCB (Dem. 11),
Which are the angles upon the other side of the base.
Therefore the angles at the base, <&c. (see Enunciation).
Which was to he shown.
CoROLLABT. — ^Henoo every equilateral triangle is also
equiangular.
Proposition 6.— Theorem.
]J two angles of a triangle he eqtial to one anotliery ths
sides (jUso which sxibtend, or a/re opposite to, tJie equal angles^
shaU he equal to one anx>thefr.
Let ABC be a triangle having the angle ABC equal to
the angle ACB.
The side AB shall be equal to the side AC.
For if AB be not equal to AC, one of them is greater Suppose
than the other. Let AB be the greater. -^ ^^^'
Construction. — From AB, the greater, cut off a part DB, Make
equal to AC, the less (L 3). DB=Aa
Join DC.
Proof. — Because in the triangles DBC, ACB, DB is
equal to AC, and BC is common to both.
Therefore the two sides DB, BC are equal
to the two sides AC, CB, each to each ;
And the angle DBC is equal to the angle
ACB (Hyp.)
Therefore the base DC is equal to the base
AB (I. 4).
And the triangle DBC is equal to the tri-
angle ACB (L 4), the less to the greater, ^^ ^^ adbc =
wUch is absurd.
Therefore AB is not unequal to AC, that is, it is equal to it.
Wherefore, if two angles, &c. Q, E, D. *
Corollary. — ^Hence every equiangular triangle is also
equilateral.
* Q. E. D. is an Abbreviation for quod erat cfenuwietrandum, t\va\i \a,
^iaikidi tffoa Uf de shoifffi or proved,"
100 G£OXETBY.
Propoeition 7.— Theorem.
Upon tlie same hose, amd on His same side ofU^ ihere eamut
he two triangles tliat have their sides, which cure terminated in
one extremitf/ of the base, egtuU to one anot/ier, and UhemM
tliose which are terminated in Hie otJier extremity.
Let the triangles AC6, AD£, upon the same base AB^
and on tlie same side of it, have, if possible.
Suppose (1 *. Their sides CA, DA, terminated in the
CA=DA. r\7\ exti-emity A of the base, equal to one
another ;
CB=:DB. / / \\ -^^ their sides CB, DB, terminated in
the extremity- B of the base, likewise
equal to one another.
Case I. — Let the vertex of each trian^
Jl il be without the other triangle.
Construction. — Join CD.
Proof. — Because AC is equal to AD (Hyp.),
The triangle ADC is an isosceles tnangle, and the an^
1 kCD - ACD is therefore equal to the angle ADC (L 6).
But the angle ACD is greater i^han the angle BCD (Ax. 9).
Therefore the angle ADC is also greater than BCD.
BDG > Much more then is the angle BDC greater than BOD.
Again, because BC is equal to BD (Hyp.),
_ The tiiangle BCD is an isosceles triangle, and the angle
z BCD." BDC is equal to the angle BCD (L 5).
But the angle BDC has been shown to be greater than the
angle BCD (Dem. 5).
z BDC = Therefore the angle BDC is both equal to, and greater than
2:"b(^. tlie same angle BCD, which is impossible.
Case II. — ^Let the vertex of one of the
triangles fall within the other.
Construction. — ^Produce AC, AD to
E and F, and join CD.
Proof. — Because AC is equal to AD
(Hyp.),
A«ain l^ NjvT, ThetriangleADCisan isosceles triangle,
^ f^ = ^ ^^ and the angles ECD, FDC, upon the other
side ofiU base CD, are ecjvia\ to oiife «£iQ>iJcist (J.. 5^
/ADC.
BDC
BCD.
/BDC =
PKOPOSITIONS. 101
Bat the angle EOD is greater than the angle BCD (Ax. 0).
Therefore ^e angle FDC is likewise greater than BCD. ^
Much more then is the angle BDC greater than BCD. c bcd.
Again, because BC is equal to BD (Hyp.),
The triangle BDC is an isosceles triangle, and the angle zbdc =
BDC is equal to the angle BCD (I. 5). ^ ^^•
But the angle BDC has been shown to be greater than the
angle BCD.
Therefore the angle BDC is both equal to, and greater than /. l bdo
the same angle BCD, which is impossible. >^°bcd.
Therefore, upon the same base, <&a Q. E* D,
Proposition 8.— Theorem.
If two triangles have two sides of the one equal to two sides
tfike other y each to ea^^h, and ha/oe likewise their bases eqtud,
the angle which is co^Uained by the two sides of the one sludl
he equal to the amgle contained by the two sides j equal to theniy
of ike other. Or,
If two triangles have three sides of the one respectively equal
to the three sides of the other, they are equal in every respect,
those angles bemg equal which are opposite to tJie equal sides.
Let ABC, DEF be two triangles which have
The two sides AB, AC equal to the two sides DE, DF, ^«« ^^
each to each, viz., AB to DE, and AC to DF, ac = df,'
And the base BC equal to the base EF. ^= sp.
Hie angle BAC shall be equal to the angle EDF.
Pboof. — ^For if the triangle ABC be applied to the triangle
DEF,
So that the point B may be on E, and the straight line BC
on EF,
The point C shall coin- a 9 O ^^^° ^^
dde TdA the point ^'fs^ \V\ ^'^.
becanse BC IB equal to EF
(Hyp.).
Therefore, BC coincid-
ing with EF, BAand AC
■hall coincide with ED
andDF.
Tar if the base BC coincidea with the base EF,
102 GEOMETRY.
But the sides BA, AC, do not coincide with the sides
ED, DF, but have a different situation, as EG, GF,
Then upon the same base, and on the same side of it, there
will be two triangles, which have their sides terminated in
one extremity of the base equal to one another, and likewise
their sides, which are terminated in the other extremity.
But this is impossible (I. 7).
•*• ^^» A^ Therefore, if the base BC coincides with the base EF, the
lycoincido sides BA, AC must coincide with the sides ED, DF.
2|5**PP Therefore the angle BAC coincides with the angle EDF,
and is equal to it (Ax. 8).
Also the triangle ABC coincides with the triangle DEF
and is therefore equal to it in every respect (Ax. 8).
Therefore, if two triangles, &c. Q. E, D.
Proposition 9.— Problem.
To bisect a given rectilineat angle, that is, to divide it into two
eqtial pa/rts.
Let BAC be the given rectilineal angle.
A It is required to bisect iti
/h Construction. — Take &ny point D in AB.
Make / \ From AO cut off AE equal to AD (I. 3).
AE= m ^Z \« Join DE.
A DEPe^ / \ / \ Upon DEj on the side remote from A, de^
quiiaterai. / \|/ \ gcribe an equilateral triangle DEF (L 1).
» JP O JoinAF.
Tlien the straight line Ai* shall bisect the a/ngle BAC.
Proof. — Because AD is equal to AE (Const), and AF is
common to the two triangles DAF, EAF;
The two sides DA, AF are equal to the two sides EA,
AF*^ each td each ;
And the base DF is eqxial to the base EF (Const.) ;
£ DAP Therefore the angle DAF is equal to the angle EAF (I. 8).
Therefore the given rectilineal angle BAC is bisected by
the straight line AF. Q. K F,
= AEAF.
Proposition 10.— Problem.
i^ Insed a given Jmite etrai^ht line^ that is, to divide it inio
two tsqrial parta^
PROPOSITIONS.
103
Let AB be the given straight lina
It is required to divide it into two equal parts.
Construction. — ^TJpon AB describe the
equilateral triangle ABC (I. 1).
Bisect the angle ACB by the straight line
CD(L9).
Then AB shoXL he cut into two equal parts
m the point D.
Pboop. — Because AC is equal to CB •'^
Hake
A ABCe-
quilateral
and
/ACDa
AD
DB.
(Const.)^ and CD common to the two triangles ACD, BCD;
The two sides AC, CD are equal to the two sides BC,
CD, each to eaoh;
And the angle ACD is equal to the angle BCD (Const.) ;
Therefore the base AD is equal to the base DB (I. 4).
Therefore the straight line AB is divided into two equal
parts in the poin€ D. Q. E. F,
Proposition 11.— Problem.
To d/ra/uo a af/raight line at right amglea to a given straight
Unefronh a given point in the same. ■
Let AB be the given straight line> and C a given point in it.
It is required to draw a straight line from the point C at
right angles to AB.
CoNSTKUOTiON. — ^Take any point D in AC.
Make CE equal to CD (L 3).
tJpon DE describe the equilateral triangle DFE (I. 1).
Join FC. y
Then FC shaU he a/t right angles
ioAB.
Proof. — ^Because DO is equal to
CE (Const.), and FC common to
the two triaingles DCF, ECF ;
The two sides DC, CF, are equal a ji
to the two sides EC, CF, each to each ;
And the base DF is equal to the base EF (Const.);
Therefore the angle DCF is equal to the angle ECF (I. 8) ; >^ DOP =
And they are adjacent angles.
But when a straight line, standing on another straight
line, makes the adjacent angles equal to one another, each of
the angles is called a right angle (Def. 10) ;
HakeCfi
s=CD an4
A DEF e-
quilatenJ.
Make
ZABEa
rights
104 GEOMETRY.
2*^F^ Therefore each of the angles DCF, EOF is a right angk
right Therefore from the given point C in the given straight line
^^^^ AB, a straight line FC has been drawn at right angles to
AB. Q, K F,
Corollary. — By help of this problem, it may be demon-
strated that
Two straight lines can/not have a common segment*
If it be possible, let the two straight lines ABC, ABD|
have the segment AB common to
both of them.
Construction. — ^From the point
B, draw BE at right angles to AB
(I- 11).
/CBE= ABC Proof. — Because ABC is a
^EBA. straight line, the angle CBE is equal to the angle EBA
and (Def. 10).
z DBE « Also, because ABD is a straight line, the angle DBE is
^ ^^^ equal to the angle EBA (Def. 10).
/. z DBE Therefore the angle DBE is equal to the angle CBE. The
= z CBE. legg to the greater; which is impossible.
Therefore two straight lines cannot have a common stg-
ment.
Proposition 12.— Problem.
To draw a straight line perpendictUa/r to a given straight
line of unlimited length, from a given point withovZ it.
Let AB be the given straight line, which may be produced
to any length both ways, and let C be a point without it.
It is required to draw from the point C, a straight line
perpendicular to AB.
Construction. — ^Take any point
D upon the other side of AB.
CD as ra- V / \ \ IVom the centre C, at the distance
dius. \ / \ >il ^^f describe the circle EGF, meet-
y ing AB in F and G (Post. 3).
Bisect FG A. p'^-^^-^ii fi Bisect FG in H (I. 10).
^^ ^ Join CF, CH, CG.
Then CH shall he perpendicular to AB.
Fboof. — ^Because FH is equal to HG (Const.), and HO
Common to the two triangle TJ'H.C, QIHG \
PROPOSITIONS.
105
Thd two 6iddS VS, HC are equal to the two sides GH,
HOy each to each ;
And the base OF is equal to the base CG (Def. 15) ;
Therefore the angle CHF is equal to the angle CHG (L 8), .'. adjacent
and they are adjacent angles. ^f!cho
But when a straight line, standing on another straight *» •^u^
line, makes the adjacent angles equal to one another, each of
the angles is called a right angle, and the straight line which
stands on the other is called a perpendicular to it ^Def. 10).
Therefore, from the given point C, a perpendicular has
been drawn to the given straight line AB. Q, JS. F,
Proposition 18.— Theorem.
UiS angles which one straight line makes with (mother upon
one side qfit, a/re either two riglU a/rtgles, or are together equal
to two right angles.
Let the straight line AB make with CD, upon one side of
it, the angles CBA, ABD.
These angles shall either be two right angles, or shall to-
gether be equal to two right angles.
Pboof. — If the angle CBA be equal to the angle ABD,
each of them is a right angle (Def. 10).
But if the angle CBA be not equal to the angle ABD,
from the point B, draw BE at right angles to CD (I. 11).
Therefore the angles CBE, EBD, are two right angles. ^|g ^
Now the angle CBE is eqiial to the two angles CBA, ABE; ^ ebd ^
to each of these equals add the angle EBD. * ^'^^ ^'
B
3(5 B 5
Therefore the angles CBijj, EBD, are equal to the three
angles CBA, ABE, EBD (Ax. 2).
Again, the angle DBA is equal to the two angles DBE,
EBA ; to each of these equals add the angle ABC.
Therefore the angles DBA, ABC, are equal to the three
angles DBE, EBA^ ABO (Ax. 2).
z £BD»
Z CBA +
Z ABE +
I EBD, al-
so Z DBA
+ Z ABO
B Z DBE
+ £> EBA
+ Z ABO.
106 OSOMETEY.
Bat the angles C6E, EBD have been Bhown to be equal
to the same three angles ;
And things which are equal to the same thing are equal
to one another ;
/. I CBE Therefore the angles CBE, EBD, are equal to the angles
= z MA I^BA, ABC (Ax. 1).
+ £ ABC. But the angles CBE, EBD are two right angles.
Therefore Qie angles DBA, ABC, are together equal to
two right angles (Ax. 1).
Therefore, the angles which one straight line, &c. Q, E. D.
Propositidn ll^Theorem.
If, at a point in a straight line, two other straight lines,
upon the opposite sides of it, make the adja>cent angles together
equal to two right ambles, these two straight lines shfUl be in
one and the sa/me straight line.
Given^ At the point B in the straight line AB, let the two
z ABD= straight lines BC, BD, upon the
M^2f^* A opposite sides of AB^ make the ad-
jacent angles ABC, ABD together
equal to twO right iangles.
BD shall be in the same straight
line with BC.
ifpossiblei / £ For if BD be not in the same
a Stm^hJ^ o 5 ''^' j> straight Ime with BC, let BE be in
line. the same straight line with it.
Proof. — ^Because CBE is a straight line, and AB meets
it in B;
Therefore the iadjacent angles ABC, ABE are together
fequal to two right angles (I. 13).
But the angles ABC, ABD, are also together equal to two
right angles (Hyp.) ;
Therefore the angles ABC, ABE, are equal to the angles
ABC, ABD (Ax. 1).
Take away the common angle ABC.
. ^ j^^ The remaining angle ABE is equal to the remaining angle
^ ^ABD, ABD (Ax. 3), the less to the greater, which is impossible ;
Therefore BE is not in tTa.e aaiofi «toa!^YflL<5k^«in^BC.
PROPOSITIONS. 107
And, in like manner, it may be demonstrated that no
other can be in the same straight line with it but BD.
Therefore BD is in the same straight line with BCl
Therefore, if at a point, &c. Q, K D.
Proposition 15.— Theorem.
1/ two straight lines ctd one (mother, the vertical, or opposite
angles shall he eqtuil.
Let the two straight lines AB, CD cut one another in the
point E.
The angle AEC shall be equal to
angle DEB, and the angle CEB to
the angle AED.
Proof. — Because the straight line "^ ^ cj^ ^
AE makes with CD, the angles CEA, z aed «
AED, these angles are together equal to two right angles tJSes.
(L 13).
Again, because the straight line DE inakes with AB the j-j.
angles AED, DEB, these also are together equal to two right z deb ==
angles (I. 13). f^
But the angles CEA, AED hard been shown to be
together equal to two Hght angles,
Therefore the angles CEA, AED are iBqual to the angles
AED^ DEB (Ax. 1).
Take away the common iangle AEli.
The remaining angle CEA is equal to the remaining angle . ^ ^^^
DEB (Ax. 3). = z DEB.
In the same manner it can be showii that the angles CEB,
AED are equal
Therefore, if twd straight lines, «fec. Q. E, Di
GoROLLART 1. — ^From this it is inanifest that if two
straight lines cut one another, the angles which they make at
the point where they cut> are together equal to four right
angles.
Corollary 2. — ^And, consequently, that all the angles
made by any ntunber of lines ineeting in 6ne point are
together equal to four right angles, provided that no on:^ Cil
the BJogJes be Included in any other angle.
108 GEOMETRY.
Proposition 16. — Theorem.
If one side of a triangle he produced, the exterior angle
shall be greater than eitlier of tlie interior opposite angles.
Let ABC be a triangle, and let its side BC be produced to D.
The exterior angle ACD shall be greater thaii either of the
j^ p interior opposite angles CBA, BAG.
Make A y^ CONSTRUCTION. — Bisect AC in E
(I. 10).
BF = BE. / ^^ J Join BE, and produce it to F,
making EF equal to BE (I. 3), and
join FC.
Proof. — Because AE is equal to
EC, and BE equal to EF (Const),
AE, EB are equal to CE, EF,
C»\ each to each ;
And the angle AEB is equal to the angle CEP, because
they are opposite vertical angles (I. 15).
Therefore the base AB is equal to the base CF (I. 4) ;
And the triangle AEB to the triangle CEF (I. 4) ;
And the remaining angles to the remaining angles, each to
each, to which the equal sides are opposite.
• zBAE Therefore the angle BAE is equal to the angle ECF
= ZBCF. (1.4).
But the angle ECD is greater than the angle ECF
(Ax. 9) ;
Therefore the angle ACD is greater than the angle BAE.
.*. zACD In the same manner, if BC be bisected, and the side AC be
^-^BAE. produced to G, it may be proved that the angle BCG (or its
equal ACD), is greater than the angle ABC.
Therefore, if one side, &c. Q, E, B.
Proposition 17.-~Theorem.
Any two angles of a tria/ngle are together less than two rigid
angles.
Let ABC be any triangle.
Any two of its angles together shall be less than two right
angles.
PBOPOSITIOKS. 109
CoNSTEUCTiON. — Produce BC to D.
Pboop. — Because ACD is the ex- /\ ^ ^
terior angle of the triangle ABC, it is
greater than the interior and opposite
angle ABC (I. 16).
To each of these add the angle Z \ ^CB~''*
ACB.
Therefore the angles ACD, ACB are greater than the i abc +
angles ABC, ACB (Ax. 4). i^^ ^
But the angles ACD, ACB are together equal to two "HfW
li^t angles (I. 13);
; Therefore the angles ABC, ACB are together less than
t two right angles.
In like manner, it may be proved that the angles BAC,
ACBy as also the angles CAB, ABC are together less than
two right angles.
Therefore, any two angles, &c. Q, E, D,
Proposition 18.— Theorem.
The greater side of every triangle is opposite tJie greater artgle.
Let ABC be a triangle, of which the side AC is greater ^q > ^^
than the side AB.
The angle ABC shall be greater than
the angle BCA.
CoNSTEUCTiON. — Because AC is / ^^--^^^^^^ ad'Lab
greater than AB, make AD equal to Z — """"^^^ ^v ,
AB (L 3), and join BD. B C
Pboof. — Because ADB is the exterior angle of the triangle ^ ^^jg >.
BDC, it is greater than the interior and opposite angle BCD ^ bcd,
(L 16). and
But the angle ADB is equal to the angle ABD ; the ^ ^^ ~
triangle BAD being isosceles (I. 5), and
Therefore the angle ABD is greater than tlie angle BCD >/bcd
(or ACB).
Much more then is the angle ABC greater than the angle
ACB.
Therefoj^^ the greater side, &c. Q, E, Z>,
no GBOMBTJiy,
Proposition 19.— Theorem,
Tlie greater angle of every triomgle is subtended hy the
greater side, or has the greater side opposite to it.
z ABC> Let ABC be a triangle, of which the angle ABC is greater
^ BCA. ^^ ^Q ajjgle BCA;
The side AC shall be greater than the side AB.
Proof. — K AC be not greater than
AB, it must either be equal to or less
than AB.
It is not equal, for then the angle
ABC would be equal to the angle BCA
(I. 5) ; but it is not (Hyp.) ;
AC not 5= Therefore AC is not equal to AB.
Neither is AC less than AB, for then the angle ABO
would be less than the angle BCA (I. 18); but it is not
(Hyp.) ;
AC not < Therefore AC is not less than AB.
And it has been proved that AC is not equal to AB ;
Therefore AC is greater than AB.
Therefore, the greater angle, (fee. Q, E, J),
Proposition SO. — ^Theorem,
Any two sides of a triangle are together greater than ihs
third side.
Let ABC be a triangle ;
Any two sides of it are together greater than the third side.
Make CONSTRUCTION. — Produce BA to the point D, making AD
AD = AC. ^^al to AC (L 3), and join DC.
Proof. — Because DA is equal to AC, the angle ADC is
equal to the angle ACD (L 5).
But the angle BCD is greater than the angle ACD (Ax. 9);
^ IS? ^ Therefore the angle BCD is greater than the angle ADC
^ ^^- jy {or BDC).
And because the angle BCD of the
triangle DCB is greater than its angle
BDC, and that the greater angle is
subtended by the greater side ;
/.DB^'BC. •** ^ Therefore the side DB is greater
ihsLn the eid^ BC (1. 19).
PROPOSITIONS. Ill
But BD is equal to BA and AC ;
Therefore BA, AC are greater than BC. ilc^BC,
In the same maimer it may be proved that AB, BC are
greater than AC ; and BC, CA greater than AB.
Therefore any two sides, <fcc. Q, K D.
Proposition 2L— Theorem.
!
If from Hie ends of Hie side of a triangle tJiere he dravm
two Hraight lines to a point vnthin the trian/gUy these shall he
fen ih4in the other two sides of the tria/ngle, but shall contain
Q greater cmgle.
Let ABC be a triangle, and from the points B, C, the ends
of the side BC, let the two straight lines BD, CD be drawii
to the point D within the triangle;
BD, DC shall be less than the sides BA, AC ;
Bat BD, DC shall contain an angle BDC greater than
the angle BAC.
CoNSTBUCTiON. — ^Produce BD to E.
Pboof. — 1. Because two sides of a
inangle are greater than the third side
(L 20), the two sides BA, AE, of the
triangle BAE are greater than BE.
To each of these add EC. _^
Therefore the sides BA, AC, are B O
greater than BE, EC (Ax. 4). S-^be +
Again, because the two sides CE, ED, of *Jit? triangle ec.
CBSD are greater than CD (I. 20),
To each of these add DB.
Therefore CE, EB are greater than CD, DB (Ax. 4).
But it has been shown that BA, AC are greater than ^^cd
BE, EC; +DB.
Much more then are BA, AC greater than BD, DC.
Pboop. — 2. Again, because the exterior angle of a
triangle is greater than the interior and opposite angle
(1 16), therefore BDC, the exterior angle of the triangle "^So
CDE, is greater than CED or CEB. ^ceb.
Pot the same reason, CEB, the exterior angle of the tri- ^ceb>
mifjk ABE, is greater Ihm the angle BAIJ or BAC, ^ ^^^
BA + AC
112
OEOICETBT.
And it has been shown that the angle BDC is greater
thanCEB;
* I BDO Much more then is the angle BDC greater than the angle
>zBAC. BAG.
Therefore, if from the ends, &c. Q. E, D,
DP.PO,
OHrespeo*
tively = A, D|
B,C.
FDas
radius.
andGH
Mndim.
FK = A.
GK=C.
Fropositioii 22^— Problem.
To make a triangle of which the sides shall he eqtuU to thm
given straight lines, hut any two whatever of these lines must
he greater than the third (L 20).
Let A, B, C be the three given straight lines, of which
any two whatever are greater than the third — ^namely, A
and B greater than C, A and C greater than B, and B and
C greater than A ;
It is required to make a triangle of which the sides shall
be equal to A, B, and C, each to each.
Construction. — ^Take a straight line DE terminated at
the point D, but imlimited to-
wards E.
Make DF equal to A, FG
equal to B, and GH equal to
C (L 3).
From the oentre F, at the
distance FD, describe the ciide
DKL (Post. 3).
From the centre G, at the distance GH, describe the
circle HLK (Post 3).
Join KF, KG.
Then the triangle KFG shaU have its sides equal to the three
straight lines A, B, C.
Proof. — Because the point F is the centre of the cirde
DKL, FD is equal to FK (Def. 16).
But FD is equal to A (Const.) ;
Therefore FK is equal to A (Ax. 1).
Again, because the pmnt G is the centre of the drde
HLK, GH is equal to GK (Def. 15).
But GH is equal to C (Const) ;
Therefore GK ia equal to (^Ajl V^^
PROPOSITIONS.
113
And FG is equal to B (Const,) ; fg
Therefore the three straight lines KF^ FG, GK are equal
to the three A, B, C, each to each.
Therefore the triangle KFG has its thi^ee sides KF, FG,
GK equal to the three given straight lines A, B, C.
Proposition 23.— Problem.
' At a given point in a given straight line, to make a recti-
Uneal angle e^uud to a given rectilineal angle.
Let AB be the given straight lino, and A the given point
in it) and DOE the given rectilineal angle.
It is required to take an angle at the point A, in the
straight line AB, equal to the rectilineal angle DC£.
GoKSTBUCTioN. — In CD, CE,
take any points D, E, and join
DE.
On AB construct a triangle
AFG, the sides of which shall be
equal to the three straight lines
CD, DE, EC — namely, AF equal D
to CD, FG to DE, and AG to EC
(I. 22) ;
Then the angle FAG slidU he equal to ilie angle DCE.
Proof. — Because DC, CE are equal to FA, AG, each to
cadi, and the base DE equal to the base FG (Const.),
The angle DCE is equal to the angle FAG (I. 8).
Therefore, at the given point A, in the given straight line
AB, the angle FAG has been made equal to the given recti*
lineal angle DCE. Q. K F,
= D
Make
A AFOso
that
AF = CD
FO = DK
AQ = CE.
Then
z DCE=s
^ FAG
Proposition 24.— Theorem.
jy two triangles have two sides of the one equal to two sides
of the other, each to each, but the angle contained by the two
tides of one of them greater than tJte angle contained by ths two
sides equal to them of the other, the base of that v:hich ka% the,
greater anple s^^i^^rea^ than th$ base of tH P*?ier,
114
GEOMETRY.
Suppose
DF>.DE.
Make z
EDG =
L BAC.
Make DO
= AC, and
= DF.
.•.BC=EG.
and
J EFG>-
- EOF.
.'.EG>EF.
Let ABC, DEF, be two triangles wliich liave
The two sides AB, AC equal to the two DE, DF, each to
each — ^namely, AB to DE,and AC to DF,
But the angle BAC greater than the angle EDF;
The base BC shall ba greater than the base EF.
CoNSTRUCTlOKT. — Let the side DF of the triangle DEF be
greater than its side DE.
A. *? Then at the point D, in
the straight line ED, make
the angle EDG equal to the
angle BAC (I. 23).
Make DG- equsd to AC
or DF (I. 3).
Join EG, GF.
Proof. — Because AB is
equal to DE (Hyp.), and AC to DG (Const.), the two sides
BA, AC are equal to the two ED, DG, each to each ;
And the angle BAC is equal to the angle EDG (Const.) ;
Therefore the base BC is equal to the base EG (I. 4).
And because DG is equal to DF (Const), the angle DFG
is equal to the angle DGF (I. 5).
But the angle DGF is greater than the angle EGF (Ax. 9);
Therefore the angle DFG is greater than the angle EGF;
Much more then is the angle EFG greater than the angle
EGF.
Ajid because the angle EFG of the triangle EFG is greater
than its angle EGF, and that the greater angle is subtended
by the greater side,
Therefore the side EG is greater than the side EF (I. 19).
But EG was proved equal to BC ;
Therefore BC is greater than EF.
Therefore, if two triangles, &c. (?, E. D.
Proposition 25.— ^Theorem.
If Ujoo triangles hwe two sides of the one equal to two sides
*ofthe other, each to each, hut the ba^e of the one greater than
the base of the other, the angle contained by the sides of that
which has the greaiefr base %hoM be greater than the am/gU
contained by the sides eg«Aal to them oj tl\A otl\.^.
PROPOSITIONS. 115
Let ABC, DEF, be two triangles, which have
The two sides AB, AC equal to the two sides DE, DF,
each to each — ^namely, AB to DE, and AC to DF,
But the bafie BC greater than the base EF ;
The angle BAC shall be greater than the angle EDF.
Peoop. — For if the angle
BAG be not greater than the ^ 1>
angle EDF, it must either be
equal to it or less.
But the angle BAC is not
equal to the angle EDF, for
then the base BC would be
equal to the base EF (L 4), but
it ia not (Hyp.);
Therefore the angle BAC is not equal to the angle EDF; £^/^p^
Neither is the angle BAC less than the angle EDF, for
then the base BC would be less than the base EF (I. 24), but
it is not (Hyp.),
Therefore the angle BAC is not less than the angle EDF. <^bdr
And it has been proved that the angle BAC is not equal
to the angle EDF;
Therefore the angle BAC is greater than the angle EDF.
Therefore, if two triangles, &c Q, E, B,
Proposition 26.— Theorem.
If two triangles, have two angles of tlie one equal to two
tmglea of tke other, each to each, and one side equal to one
side — namelg^ eitlisr the side adjacent to tlie equal angles in
each, or the side opposite to them ; then sludl tlie other sides he
equal, each to eajch; and also the third a/ngh of the one equal
to the third angle of the other. Or,
If two angles arid a side in one triangle he respectively equal
to two angles and a corresponding side in a/nother triangle, tlie
triangles slwU he equal in every respect.
Let ABC, DEF be two triaiigles, which have ^
The angles ABC, BCA equal to the angles DEF, EMJ;
each to each— namely, ABC to DEF, and BCA to EFD;
Also one side equal to one side.
Case 1. — ^Firot, let the sides adjacent to the equal angles bcSep.
in each be equal— namely, BC to EF;
IIG
GEOMETRY,
Suppose
AB>DE.
Make
BO = DE.
/. I GCB
= Z DFE.
ABnot
unequal
to
lequa
Snppofld
BC>EF
Hake
Then shall the side AB be equal to DE, the side AC to
DP, and the angle BAG to the angle EDF.
For if AB be not equal to DE, one of them must be
ij) greater than the other. Let AB
be the greater of the two.
Construction. — ^Make BG
equal to DE (I. 3), and join GC.
Proof. — Because BG is
equal to DE (Const), and BC
1? is equal to EF (Hyp.), the two
sides GB, BC are equal to the two sides DE, EF, each to
each.
And the angle GBC is equal to the angle DEF (Hyp.);
Therefore the base GC is equal to the base DF (I. 4),
And the triangle GBC to the triangle DEF (I. 4),
And the other angles to the other angles, each to each, to
which the equal sides are opposite;
Therefore the angle GCB is equal to the angle DFE (I. 4).
But the angle DFE is equal to the angle BCA (Hyp.) ;
Therefore tibe angle GCB is equal to the angle BCA (Ax.
1), the less to the greater, which is impossible;
Therefore AB is not unequal to DE, that is, it is equal to
it; and BC is equal to EF (Hyp.) ;
Therefore the two sides AB, BC are equal to the two sides
DE, EF, each to each,
And liie angle ABC is equal to the angle DEF (Hyp.) ;
TTierefore the base AC is equal to the base DF (1. 4),
And the third angle BAC to the third angle EDF (I. 4).
Case 2. — Next, let the sides which affe opposite to the
equal angles in each triangle be equal to one another — ^namely,
AB equal to DE.
Likewise in this case the other sides shall be equal, AC to
A I) DF,andBCtoEF;andalsothe
angle BAC to the angle EDF.
For if BC be not equal to
EF, one of them must be greater
than the other. Let BC be th^
greater of the two.
C0N3TRUCTIQNt-T^Make BH
eqml to EF (I. 3), ftxxd Jgiu AH,
PROPOSITIONS. 117
Proof. — Because BH is equal to EF (Const.), and AB is bh = ep,
equal to DE (Hyp.), the two sides AB, BH are equal to the ~ ^^
two sides DE, EF, each to each.
And the angle ABH is equal to the angle DEF (Hyp.) ; j j^ =
Therefore the base AH is equal to the base DF (I. i),
And the triangle ABH to the triangle DEF ^I. 4),
And the other angles to the other angles, each to each, to
which the sides are opposite ;
Therefore the angle BHA is equal to the angle EFD (I. 4). . ^ ^jj^
But the angle EFD is equal to the angle BC A (Hyp.) ; = z efd
Therefore tibe angle BHA is also equal to the angle BCA "^ ^^^
(Ar.1);
That is, the exterior angle BHA of the triangle AHC, is
equal to its interior and opposite angle BCA, which is
impossible (I. 16) ;
Therefore BC is not unequal to EF — that is, it is equal to it ; BC not
and AB is equal to DE (Hyp.) ; ^ ^ S\
Therefore the two sides AB, BC are equal to the two sides
DE, EF, each to each.
And the angle ABC is equal to the angle DEF (Hyp.) ;
Therefore the base AC is, equal to the base DF (I. 4),
And the third angle BAC is equal to the third angle
£DF (I. 4).
Therefore, if two triangles, <fec, Q. E, D,
Proposition 27.— Theorem. '
7/* a straight line falling upon two other straight lines make
the alternate a/ngles equal to one another, these two straight
lines shall be parallel.
Let the straight line EF, which
faUs upon the two straight lines
AB, CD, make the alternate angles ^ J'V B Given
AEF, EFD, equal to one another. /^ \.r. i efd
AB shall be parallel to CD.
For if AB and CD be not parallel,
they will meet if produced, either
towards B, D, or towaixis A, C.
Let them be produced^ and meet towards B, D, m \)aa
jx>int G,
118 GEOMETKY.
z AEF> Proof. — ^Then GEF is a triangle, and its exterior angle
z EFG, AEF is greater than the interior and opposite angle EFG
and also (I. 16).
= z EFG. But the angle AEF is also equal to EFG (Hyp.), which is
impossible;
Therefore AB and CD, being produced, do not meet
towards B, D.
In like manner it may be sKown that they do not meet
towards A, C.
But those straight lines in the same plane which being
produced ever so far both ways do not meet are parallel
(Def. 34) ;
Therefore AB is parallel to CB.
Therefore, if a straight line, &c, Q. E, JX
Proposition 28.— Theorem.
If a straight linefaUing upon two oilwr atraigJU lines make
the exterior angle eqiuU to the interior a/nd opposite upon the
earne side of the line, or maJce the interior angles upon the same
side togeifier equal to two right angles, the two straight Unas
sliaU he parallel to one another.
Let the straight line EF, which falls upon the two straight
lines AB, CD, make —
The exterior angle EGB equal to the interior and opposite
E\ angle GHD, upon the same side 3
Or make the interior angles on the
^ ^"^^ B same side, the angles BGH, GHD,
together equal to two right angles ;
-D AB shall be parallel to CD.
Proof 1. — Because the angle EGB
^^ is equal to the angle GHD (Hyp.),
And the angle EGB is equal to the angle AGH (I. 16) ;
c AGH=: Therefore the angle AGH is equal to the angle GHD
' "^^ (Ax. 1), and these angles are alternate;
Therefore AB is parallel to CD (I. 27).
Proof 2. — ^Again, because the angles BGH, GHD are
equal to two right angles (Hyp.),
And the angles BGH, AGH are also equal to two right
angles (L 13).
L GHD.
PKOPOSITIONS. 119
Therefore the angles BGH, AGH are equal to the angles /. ^ boh
BGH, GHD (Ax. 1). 1;^ ^g
Take away the common angle BGH. + ^ ®*"^-
Therefore the remaining angle AGH is equal to the remain- .-. ^ agh
ing angle GHD (Ax. 3), and they are alternate angles. = ^ ^^^
Therefore AB is paraUel to CD (I. 27).
Therefore^ if a straight line, &c. Q, E. D.
Proposition 29.— Theorem.
If a straight linefaU v/pon two parallel aPraightlineSyit maJces
IKr aUerruUe angles equal to one anotlier^ amd the exterior angle
equal to Hie interior amd opposite upon t/ie same side; and
also t/ie two interior angles upon t/ie sa/rne side together equal to
two right angles.
Let the straight line EP fall upon the parallel straight
lines AB, CD;
The alternate angles AGH, GHD shall be equal to one
another.
The exterior angle EGB shall be
equal to GHD, the interior and op-
posite angle upon the same side ; A ^^^ B
And the two interior angles on the
same side BGH, GHD shall be to- c-
gether equal to two right angles.
For if AGH be not equal to ^^ z agii >
GHD, one of them must be greater ghd.
than the other. Let AGH be the greater. (suppose.)
Pboop. — Then the angle AGH is greater than the angle
GHD; to each of them add the angle BGH.
Therefore the angles BGH, AGH are greater than tlio
angles BGH, GHD (Ax. 4).
Bat the angles BGH, AGH are together equal to two right
angles (1. 3). . ^ ^^^
Therefore the angles BGH, GHD are less than two right + z ghd
, ® ' -<:two right
angles. angles.^
But if a straight line meet two straight lines, so as to
make the two interior angles on the same side of it taken
together less than two right angles, these 8traig\it\m<&^\>^m*^
120 GEOMETRY.
continually produced, shall at length meet on that side on
which are the angles which are less than two right angles
(Ax. 12);
Hence Therefore the straight lines AB, CD will meet if produced
and are ' £ut they cannot meet, because they are parallel straight
•«»^- lines (Hyp.);
not TO©. Therefore the angle AGH is not imequal to the angle
<i«^ to GHD — that is, it is equal to it.
But the angle AGH is equal to the angle EGB (I. 15) ;
and Therefore the angle EGB is equal to the angle GHD
Add to each of these the angle BGH.
Therefore the angles EGB, BGH, are equal to the angles
BGH, GHD (Ax. 2).
But the angles EGB, BGH, are equal to two right angles
(I. 13).
ZBOH + Therefore also BGH, GHD, are equal to two right angles
zGHD= (Ax. 1).
JSS^iSf ' Therefore, if a straight line, &c. Q, E. D.
/^^"^^ , Proposition 80.— Theorem.
^ '-''^'/iv ^Straight tinea which are parallel to the same straiglU lines
^y^\ ■<^\' are paralld to one a/nother.
Let AB, CD be each of them parallel to EF ;
AB shall be parallel to CD.
Construction. — Let the straight line GHK cut AB,
. EF, CD.
^ ^g^?! / Proof. — Because GHK cuts thepar-
i GHF,"" A ^ B allel straight lines AB, EF, the angle
/ AGH isequal to the angle GHF (L 29).
T OHP = * / * Again, because GK cuts the parallel
z GKD. K/ jj straight lines EF, CD, the angle GHF
^ T is equal to the angle GKD (L 29).
/ And it was shown that the angle
AGK is equal to the angle GHF ;
_/. z AGK Therefore the angle AGK is equal to the angle GKD
"^ ' (Ax. 1), and they are alt^ma^ wa^lea \
PROPOSITION. 121
Therefore AB is parallel to CD (I. 27).
Therefore, straight lines, kc Q, E. I),
Proposition 81. — Problem.
To drwvo a straight line through a given poirUy parallel to a
given straight line.
Let A be the given point, and BC the given straight line.
It is required to draw a straight line through the point A,
parallel to BC.
Construction. — ^Tn BC take any B -^ ^
point D, and join AD. /
At the point A, in the straight line /
AD, make the angle DAE equal to the S i> 2 Make
angle ADC (I. 23). i XSS."
Produce the straight line EA to F.
nm EF shaU be parallel to BC.
Proof. — Because the straight line AD, which meets the ''Jjey ^^
two straight lines BC, EF, makes the alternate angles E AD, angles.
ADC equal to one another ;
Therefore EF is parallel to BC (I. 27).
Therefore, the straight line EAF is drawn through the
given point A, parallel to the given straight line BC. Q, E, F.
Proposition 82.— Theorem.
If a side of any triangle he produced ^ tlie exterior atigle is
equal to the two interior and opposite angles ; and tlie three
interior angles qf every triangle are eqv/al to two right angles.
Let ABC be a triangle, and let one of its sides BC be pro-
duced to D;
The exterior angle ACD shall be equal to the two interior
and opposite angles CAB, ABC ; ^
And the three interior angles y\
of the triangle — namely, ABC,
BCA, CAB, shall be equal to
two right angles.
Construction. — ^Through the ^ ' Make
point a draw CE pai-aUel to AB (I. 31). ^£1^^'
122 GEOMETRY.
Then Proop. — Becaiise AB is parallel to CE, and AC meets
z ACE ~ them, the alternate angles BAG, ACE are equal (L 29).
j^^i ' Again, because AB is parallel to CE, and BD falls upon
z. ECD = them, the exterior angle ECD is equal to the interior and
z ABC. opposite angle ABC (I. 29).
But the angle ACE was shown to be equal to the angle
BAC;
'— ^ BA? Therefore the whole exterior angle ACD is equal to the
+ L ABC. two interior and opposite angles BAC, ABC (Ax. 2).
Add To each of these equals add the angle ACB.
Therefore the angles ACD, ACB are equal to the three
angles CBA, BAC, ACB (Ax. 2).
But the angles ACD, ACB are equal to two right angles
(I. 13) ;
Therefore also the angles CBA, BAC, ACB are equal to
two right angles (Ax. 1).
Therefore, if a side of any triangle, &c. Q, E, D,
CorollaTeiy 1. — All tJie interior angles of any rectilineal
figure, togetlier with four right anglea, are equal to ttvice as
many right angles as tlie figure lias sides.
For any rectilineal figure ABCDE can, .by drawing
straight lines from a point E within the figure to each angle,
be divided into as many triangles as the
figure has sides.
And, by the preceding proposition, the
angles of each triangle are equal to two
right angles.
_ Therefore all the angles of the triangles
^ are equal to twice as many right angles
as there are triangles; that is, as there are sides of the
figure.
But the same angles are equal to the angles of the figure,
together with the angles at the point F ;
And the angles at the point F, which is the common
vertex of all the triangles, are equal to four right angles (L
15, Cor. 2) ;
Therefore all the angles of the figure, together with four
right angles, are equal to twice as many right angles as the
JSgwre haa sides.
PROPOSITIONS. 123
CoROLLABY 2. — All the exterior angles of cmy rectilineal
figure are together equal tofowr right angles.
The interior angle ABC, with its adjacent exterior angle
ABD, is equal to two right angles (I. 1 3) ;
Therefore all the interior, together with all the exterior
angles of the figure, are equal to twice as many right angles
as the figure has ^ides.
But all the interior angles, together
vith four right angles, are equal to
twice as many right angles as the
figure has sides (I. 32, Cor. 1) ;
Therefore all the interior angles^
together with all the exterior angles,
are equal to all the interior angles
and four right angles (Ax. 1).
Take away the interior angles which are common ;
Therefore all the exterior angles are equal to four right
angles (Ax. 3).
Proposition S3.— Theorem.
The straight lines which join tlie extremities of two equal
and paraUd straight lines towards the sanie iJO/rts a/re also
themselves equal cmd parallel.
Let AB and CD be equal and parallel straight lines joined
towards the same parts by the straight lines AC and BD ;
AC and BD shall be equal and parallel.
Construction. — Join BC.
Proof. — ^Because AB is parallel to CD, and BC meets ^ ^_^
them, the alternate angles ABC, BCD are equal (I. 29).
Because AB is equal to CD, and BC common to the two
triangles ABC, DCB, the two sides A
AB, BC are equal to the two sides
DC, CB, each to each ; *'
And the angle ABC was proved
to be equal to the angle BCD ;
Therdfore the base AC is equal to the base BD (I. 4), ^^^ =
And the triangle ABC is equal to the triangle BCD (I. 4), '
And the other angles are equal to the other angles, each to
each, to which the equal aides axe opposite ;
124 QEOMETRT.
*" ACB = Therefore the angle ACB is equal to the angle CBD.
L CBD." And because the straight line BC meets the two straight
lines AC, BD, and makes the alternate angles ACB, CBD*
equal to one another ;
Therefore AC is parallel to BD (I. 27) ; and it was shown
to be equal to it.
Therefore, the straight lines, &c. Q, E, D.
*
Proposition 34. — Theorem.
Tlie opposite sides and angles of a paraUehgram are equal
to one another, and tlie diagonal bisects the parcUlelogra/m — t/uU
is, divides it into two equal parts.
Let ACDB be a parallelogram, of which BC is a diagonal ;
The opposite sides and angles of the figure shall be equal
to one another.
And the diagonal BC shall bisect it.
Proof. — Because AB is paraUel to CD, and BC meets
^ ABC = them, the alternate angles ABC, BCD
' A B are equal to one another (I. 29) ;
\ ^y^^\^ Because AC is parallel to BD, and
^^ \ ^y^ \ BC meets them, the alternate angles
z ACB= \^ \ ACB, CBD are equal to one ano&er
. CBD. ^ t ^ 29) ;
Therefore the two triangles ABC, BCD have two angles,
ABC, BCA in the one, equal to two angles, BCD, CBD in
the other, each to each ; and the side BC, adjacent to the
equal angles in each, is common to both triangles.
Therefore the other sides are equal, each to each, and the
third angle of the one to the third angle of the other —
.'. AD = namely, AB equal to CD, AC to BD, and the angle BAC to
bS; ^bac the angle CDB (I. 26).
= z CDB, ^n(i because the angle ABC is equal to the' angle BCD,
and the angle CBD to the angle ACB,
and Therefore the whole angle ABD is equal to the whole
i^icS. angleACD(Ax.2).
And the angle BAC has been shown to be equal to the
angle BDC; therefore the opposite sides and angles of a
parallelogram are equal to one another.
Also the diagonal bisects it.
PROPOSITIONS.
125
For AB being equal to CD, and BC common,
The two sides Afe, BC are equal to the two sides CD and
CB, each to each.
And the angle ABC has been shown to be equal to the
angle BCD ;
Therefore the triangle ABC is equal to the triangle BCD ^*xbc -
(I. 4), A BCD."
* And the diagonal BC divides the parallelogram ABCD
: into two equal parts.
Therefore, the opposite sides, &c, Q, E, i>.
Froposition 36.— Theorem.
PivrdUdograTM vpon the same base, and between the same
parallels, are eqiud to one another, ,. ^
Let the parallelograms ABCD, EBCF be on the same base
BC, and between the same parallels AF, BC;
The parallelogram ABCD shall be equal to the parallelo-
gram EBCF.
Case 1.— K the sides AD, DF of the ^
parallelograms ABCD, DBCF, opposite
to the l^use BC, be terminated in the
aame point D, it is plain that each of the
parallelograms is double of the triangle
DBG (I. 34), and that they are therefore equal to one
mother (Ax. 6).
Case 2. — But if the sides AD, £F, opposite to the base
BC, of the parallelograms ABCD, EBCF, be not terminated
in the same point, then — A liia w A E d
Pboof. — Because \ )[ / \
ABCD is a parallelo- \ /\ / \
gram, AD is equal to \/ \/
BC(I. 34). B ^
For the same reason EF is equal to BC; ef = bc.
Therefore AD is equal to EF (Ax. 1), and DE is common;
Therefore the whole, or the remainder, AE, is eqi^al to the
whole, or the remainder, DF (Ax. 2, or 3), .\AE=df,
And AB is equal to DC (I. 34).
Therefore the two EA, AB are equal to tfre two FD, DC*
each to ^ach;
AD = Ba
Hence
126 GEOMETRY.
And the exterior angle FDC is equal to the interior EAB
(I. 29);
Therefore the base EB is equal to the base FC (L 4),
a'eab = And the triangle EAB equal to the triangle FDC (L 4).
A FDC. Taj^e ^Q triangle FDC from the trapezium ABCF, and
from the same trapezium ABCF, take the triangle EAB, and
the remainders are equal (Ax. 3),
That is, the parallelogram ABCD is equal to the parallelo-
gram EBCF.
Therefore, parallelograms, &c. Q. E. D,
Proposition 36.— Theorem.
Pa/raUdograms upon equaZ bases, and between tlie same
parallels, are equal to OTie another.
Let ABCD, EFGH be parallelograms on equal bases BC,
FG, and between the same parallels AH, BG;
The parallelogram ABCD shall be equal to the {Parallelo-
gram EFGH.
CoNSTEUCTioN. — JoinBE, CH.
Proof. — Because BC is equal
to FG (Hyp.), and FG to EH
(I. 34),
BC = EH, [ x^ U^ \ I Therefore BC is equal to EH
^ (Ax. 1) ; and they are parallels,
and joined towards the same parts by the straight lines
BE, CH.
But straight lines which join the extremities of equal and
parallel straight lines towai^ds the same parts, are themselyes
equal and parallel (I. 33) ;
iTE^rr CH. Therefore BE, CH are both equal and parallel;
EBCH a Therefore EBCH is a parallelogram (De£ 35),
paraiieio- And it is equal to the parallelogram ABCD, because they
equS'each ^^^ ^^ the Same base BC, and between the same parallels BC,
"fthe AH (I. 35).
For thfc like reason, the parallelogram EFGH is equal to
the same parallelogram EBCH;
Therefore the parallelogram ABCD is equal to the parallelo-
gram EFGH (Ax. 1).
Ijberefore^ parallelograms, ^. Q. E. D«
PROPOSITIONS. 127
Proposition 87. — Theorem.
Tricmgles upo7i tlie same hose, and between tlie same parallels,
are eqtuil to one another.
Let the triangles ABC, DBC be on the same base EC, and
between the same parallels AD, BO ;
The triangle ABO shall be equal to the triangle DBO.
Construction. — Produce AD both ways, to the points E, F.
Through B draw BE parallel to CA, and through draw
CF parallel to BD (I. 31).
Proof. — Then each of the E a. p p Fijfiirea
figures EBCA, DBCr, is a parol. V K7V 7 gBCA.mi
lelogram (Def. 35), and they are \ I /\\ / equ^^i
equal to one another, because the^
are on the same base BO, and
between the same parallels BO, EF
(L35.);
And the triangle ABC is half of the parallelogram EBCA, an«i th©
because the diagonal AB bisects it (I. 34) ; are^respec-
And the triangle DBC is half of the pamllelogram DBCF, ^^f^^^
because the diagonal DO bisects it (I. 34).
But the halves of equal things are equal (Ax. 7) ;
Therefore the triangle ABO is equal to the triangle DBC,
Therefore, triangles, &c. Q. J^. i>.
Proposition 88.— Theorem.
Triangles vpon equal bases, and behueen tlie same parallels,
are equal to one another.
Let the triangles ABC, DEF, be on equal bases BC, EF,
and between the same parallels BF, AD.
The triangle ABC shall be equal to the triangle DEF.
Construction. — Produce AD both ways to the points
G,H.
Through B draw BG par- » A P H
allel to CA, and through F
draw FH parallel to ED
(L 31). \ / \ I \ I Figures
Proof.— Then each of the \/ \ / \f ^ V""^^^
figures GBOA. DEFH, is a ^ C £ ^ ^^^^
128 GEOMETRY.
parallelogram (Def. 35), and they are equal to one another, be-
cause they are on equal bases BC, EF, and between the
same parallels BF, GH (I. 36);
and the And the triangle ABO is half of the parallelogram
Sl'SSfof ^SCA, because the diagonal AB bisects it (I. 34) ;
these re- And the triangle D£F is half of the parallelogram
spectiveiy. j)EFH, because the diagonal DF bisects it (I. 34).
But the halves of equal things are equal (Ax. 7);
Therefore the triangle ABO is equal to the triAHgle DEF.
Therefore, triangles, &c, Q, E. D,
Proposition 89.— Theorem*
Equal triangles vpon tlie same hose, cmd on the same side
qfity are between the same pa/rallels.
Let the equal triangles ABO, DBO be upon the same base
BO, and on the same side of it ;
They shall be between the same parallels.
Construction. — Join AD ; AD shall be parallel to BO.
t^BC^^ A.^ p For if it is not, through A draw AE
pose. ^"x^^ — K^Kv parallel to BO (I. 31), and join EC.
Proof. — The triangle ABC is equal to
the triangle EBC, because they are upon
the same base BO, and between the same
B o parallels BO, AE (I. 37).
But the triangle ABO is equal to the triangle DBO (Hyp.);
a'dbc = Therefore the triangle DBC is equal to the triangle EBC
aba^Sty!* (-^* ^)f *^® greater equal to the less, which is impossible ;
Therefore AE is not parallel to BC.
In the same manner, it can be demonstrated that no line
passing through A can be parallel to BO, except AD ;
Therefore AD is parallel to BO.
Therefore, equal triangles, &c. Q, E. D,
Proposition 40.— Theorem.
Equal triangles upon the same side of equal bases, thai
are in the sa/me straight line, are between the same parallels,
Let the equal triangles ABO, DEF, be upon the same side
of equal bajses BO, EF, in t»he B»?ae a^^raight line BF,
Then
PROPOSITIONS.
129
The triangles ABC, DEF shall be between the same
parallels.
Construction.— Join AD ; AD shall be parallel to BP.
For if it is not, through A draw AG paraUel to BF (I. 31), titTSr
and join GF. tuppoM.
Proof.— The triangle ABC is equal to the triangle GEF,
because they are upkm equal bases BC, EF, and are between
the same parallels BF, AG
(L38).
But the triangle ABC is
equal to the triangle DEF ;
Therefore the triangle DEF
b equal to the triangle GEF
(Ax. 1), the greater equal to
the less, which is impossible ;
Therefore AG is not parallel to BF.
In the same maimer, it can be demonstrated that no line,
passing Jihrough A, can be parallel to BF, except AD ;
Therefore AD is parallel to BF.
Uierefore, equal triangles, <&c.
A DEP =
A GEF, aa
absurdity.
Proposition 4L— Theorem.
If a parallelogram and a tria/ngle he upon the same hose,
md between tlie same parallels, the parallelogram shall he
doMe of the triangle.
Let the parallelogram ABCD, and the triangle EBC be
upon the same base BC, and between the same parallels
BC,AE;
The parallelogram ABCD shall be double h 1> b
ofthe triangle EBC.
Construction. — Join AC.
Proof. — ^The triangle ABC is equal to
the triangle EBC, b^use they are upon
the same base BC, and between the same
parallels BC, AE (I. 37).
But the parallelogram ABCD is double of the triangle And porai-
ABC, because the diagonal AC bisects the parallelogram (I. 2'*^^;^
A ABO =
A EBC.
130 GEOMETRY,
•Therefore the parallelogram ABCD is also double of the
triangle EBC (Ax. 1).
Therefore, if a paraUelogram, &c. Q, E, D,
Proposition 42.— Problem.
To describe a pa/ralMogram tJuit sliaU he equal to a given
triangle, and have one of its angles eqtuil to a given rectilineal
angle.
Let ABC be the giyen triangle, and D the given recti-
lineal angle ;
It is required to describe a parallelogram that shall be
equal to the given triangle ABC, and have one of its angles
equal to D.
Make BE A F G CONSTRUCTION. — Bisect BC in E (I.
- EC A-? 7 10), and join AE. ^
/ \ I I -^t *^® point E, in the straight lino
?CEP=D / f\ I m CE, make the angle CEP equal to D
(I. 23).
Through A draw AFG parallel to
EC (I. 31).
Through C draw CG parallel to EF (I. 31).
Then FECG is the parallelogrami required.
Proof. — Because BE is equal to EC (Const.), the trian^e
ABE is equal to the triangle AEC, since they are upon equal
bases and between the same parallels (I. 38) ;
^A^A^ Therefore the triangle AJBC is double of the triangle
and also AEC.
S^G - But the parallelogram FECG is also double of the triangle
2 A ABC. AEC, because they are upon the same base, and between fiie
same parallels (I. 41) ;
Therefore the parallelogram FECG is equal to the triangle
ABC (Ax. 6),
And it has one of its angles CEF equal to the given angle
D (Const.).
Therefore a parallelogram FECG has been described equal
to the given triangle ABC, and having one of its angles CEF
^ual to the given an^leD. Q, E, F,
r»oposiTioNs. 131
Proposition 43.— Theorem.
TJie complements of t/is 2^curo>llelogra7n8 lohich are about Vie
diagonal of any paraUehgra/ni are eqtud to one a/notlier.
Let ABCD be a parallelogram, of which the diagonal is
AC ; and EH, GF parallelograms about AC, that is, through
which AC passes ; and £K, KD the other parallelograms,
which make up tiie whole figure ABCD, and are therefore
called the complements.
The complement BK shall be equal to the complement
KD.
Pboop, — Because ABCD is a parallelogram, and AC its .
diagonal, the triangle ABC is equal to the triangle ADC a adc/
(L 34).
Again, because AEKH is a paral- ^- — S T>
lelogram, and AK its diagonal, the j. / Nj K j aiw
triangle AEK is equal to the triangle I /x r a aek
AJml(L34). I l\ ^
For the like reason the triangle / / x. / ^^^
KGC is equal to the triangle KFC. / / \/ a kgc
Therefore, because the triangle ^ ^ c a kfc.
AEK is equal to the triangle AHK, and the triangle KGC
to KFC,
The triangles AEK, KGC are equal to the triangles
AHK, KFC (Ax. 2).
But the whole triangle ABC was proved equal to the
whole triangle ADC ;
Therefore the remaining complement BK is equal to the .*. bk=:
remaining complement KD (Ax. 3).
Therefore, the complements, &c. Q. K D.
Proposition 44.— Problem.
To a given straight line to a/pply a parallelograrrii which
shaU he equal to a given triangle, amd Jmve one of its angles
equal to a given rectilineal a/ngle.
Let AB be the given straight line, C the gtveti tnasw^^^^
and D the given angle.
132
GEOMETBT.
Make
parallelo-
irram
BEFG =
It is required to apply to the straight line AB a parallelo
gram equal to the triangle C, and having an angle equal
toD.
CoNSTBUCTiON 1. — Make the parallelogram BEFG equal
_ to the triangle C, and having the angle ££G equal to the
r^aii a^gle D (L 42) ;
^ D and '^^ ^®* ^® parallelogram BEFG be made so that BE may
EBAa be in the same straight line with AB.
g^«**' Produce FG to H.
Through A draw AH parallel to BG or EF (I. 31).
Join HB.
Proof 1. — Because the straight line HF falls on the
parallels AH, EF, the angles AHF, HFE are together equal
to two right angles (I. 29).
Therefore the angles BHF, HFE are together less than
two right angles (Ax. 9). But straight lines which with
another straight line make the interior angles on the same
side together less than two right angles, will meet on that
side, if produced far enough (Ax. 12) ;
HBandFE Therefore HB and FE shall meet if produced.
Construction 2. — Produce HB and FE towards BE,
and let them meet in K.
F^rea
LB = BF,
Through K draw KL parallel to EA or FH (I. 31).
Produce HA, GB to the points L, M.
Then LB shall he the parallelogram required,
Pboop 2. — Because HLKF is a parallelogram, of which
the diagonal is HK ; and AG, ME are the parallelograms
about HK ; and LB, BF are the complements ;
Therefore the complement LB is equal to the complement
BF (I. 43).
PROPOSITIONS. 133
But BF is equal to the triangle C (Const.) ; But_
Therefore LB is equal to the triangle C (Ax. 1). .-. lb= '
And because the angle GBE is equal to the angle ABM ^^'*
(L 15), and likewise to the angle D (Const.) ;
Therefore the angle ABM is equal to the angle D (Ax. 1). ^^^jsa =
Therefore, the parallelogram LB is applied to the straight z obe s
line AB, and is equal to the triangle C, and has the angle ^ ^*
ABM equal to the angle D. Q. E, F.
Proposition 45.— Problem.
To describe a pa/raUelogram equal to a given rectilineal
figure^ cmd liaving an aTujle equal to a given rectilineal angle.
Let A BCD be the given rectilineal figure, and E the
given rectilineal angle.
It is required to describe a parallelogram equal to ABCD,
and having an angle equal to E.
Construction. — Join
DB. A B y <v y
Describe the parallelo- \ \ I «i*
gram FH equal to the \ \ k // =*a adS
triangle ADB, and \ / \ / / / offoM^
having the angle FKH \ / \ / / / a dbg,
equal to the angle E \ \ z*ohm =
(1.42). \ \ k — r-Ti ^E.
To the straight line
GH apply the parallelogram GM equal to the triangle DBC,
and having the angle OHM equal to the angle E (I. 44).
Then thejigvre FKML sJiaU he the paraUehgram required.
Proof. — Because the angle E is equal to each of the angles
FKH, OHM (Const),
Therefore the angle FKH is equal to the angle GHM
(Ax. 1).
Add to each of these equals the angle KHG ;
Therefore the angles FKH, KHG are equal to the angles
KHG, GHM (Ax. 2).
But FKH, KHG are equal to two right angles (L 29);
Therefore also KHG, GHM are equal to two right angiles
(Ax. 1).
Then
134 GEOMETRY.
And because at the point H, in the straight line GH, the
two straight lines KH, HM, on the opposite sides of it, make
the adjacent angles together equal to two right angles,
KHM is a Therefore KH is in the same straight line with HM (1. 14).
^ight ^j^^ because the straight line HG meets the parallels
KM, FG, the alternate angles MHG, HGF are equal (I. 29).
Add to each of these equals the angle HGL;
Therefore the angles MHG, HGL are equal to the angles
HGF, HGL (Ax. 2).
But the angles MHG, HGL are equal to two right angles
(I. 29);
Therefore also the angles HGF, HGL are equal to two
^^^ right angles,
FGL is a And tiierefore FG is in the same straight line with GL
^^^ (1.14). ^ . ,
And because KF is parallel to HG, and HG parallel to
ML (Const.);
Therefore KF is parallel to ML (I. 30).
And KM, FL are parallels (Const);
.*. KPLMa Therefore KFLM is a parallelogram (Def. 35).
gjraMeio- j^^ because the triangle ABD is equal to the parallelo-
gram HF, and the triangle DBC equal to the parallelogram
GM (Const.),
And the Therefore the whole rectilineal figure ABCD is equal to
5^ ig the whole parallelogram KFLM (Ax. 2).
equal to it. Therefore, the parallelogram KFLM has been described
equal to the given rectilineal figure ABCD, and having the
angle FKM equal to the given angle E. Q, E, F.
Corollary. — From this it is manifest how to apply to a
given straight line a parallelogram, which shall have an angle
equal to a given rectilineal angle, and shall be equal to.a given
rectilineal figure— namely, by applying to the given straight
line a parallelogram equal to ^e first triangle ABD, and
having an angle equal to the given angle ; and so on (I. 44).
Proposition 46.— Problem.
To describe a square upon a given straight line.
Let AB be the given straight line;
It 2» required to describe a Bc\vvax^ w^n AB.
PROPOsiTioifs. 136
CoNSTBUcrriON. — ^From the point A draw AC at right
angles to AB (I. 11),
And make AD equal to AB (I. 3).
Through the point D draw DE parallel
to AB (I. 31).
Through the point B draw BE parallel
to AD (I. 31).
Hhein, ADEB bIwJI be the stpmre re-
quired.
Pboof. — Because DE is parallel to
AB, and BE pai-allel to AD (Const.), aI Ib ^feto-
therefore ADEB is a parallelogram; itmil
Therefore AB is equal to DE, and AD to BE (I. 34).
Bat ATt is equal to AD (Const);
Therefore the four straight lines BA, AD, DE, EB are
equal to one another (Ax. 1),
And the parallelogram ADEB is therefore equilateral Stwd^"*
likewise all its angles are right angles.
For since the straight line AD meets the parallels AB,
DE, the angles BAD, ADE are together equal to two right
angles CL. 29).
But BAD is a right angle (Const.) ;
Therefore also ADE is a right angle (Ak. 3).
But the opposite angles of parallelograms are equal (I; 34) ;
Therefore each of the opposite angles ABE^ BED is a right
angle (Ax. 1) ;
Therefore the figure ADEB is rectangular ; and it has i* " rec-
been proved to be equilateral; therefore it is a square ^"'
(Def.30).
Therefore, the figure ADEB is a squaite, and it is described /.a square.
upon the given straight line AB. Q, E, Fi
CoBOLLARY. — llence every parallelogram that has onQ
right angle has all its angles right angles^
Proposition 47.— TheSrein.
In any riglU^iiigled tria7igte, tJie square whicJb is described
upon the side opposite to tJie right angle is equal to tlie sqicarea
described upon the sides which coiitain tlie right arigle.
136
GEOMETRY.
CAG is a
straight
line.
BAH is a
straight
line.
A ABD =
A FBC.
Hence
]>arallelo-
gram BL
= square
GB, and
parallelo-
gram CL
Let ABC be a right-angled triangle, having the right
angle BAC ;
The square described upon the
side BO shall be equal to the
squares described upon BA, AC.
OoNSTRuerioif. — On BO de-
scribe the square BDEC (I. 46).
On BA, AC describe the squares
GB, HC (I. 46).
Through A draw AL parallel
to BD or CE (I. 31).
Join AD, FC.
Proof. — Because the angle BAC
is a right angle (Hyp.), and that the angle BAG is also a
right angle (Def. 30),
The two straight lines AC, AG, upon opposite sides of
AB, make with it at the point A the adjacent angles equal
to two right angles ;
Therefore CA is in the same straight line with AG (1. 14).
For the same reason, AB and AH are in the same straight
line.
Kow the angle DBC is equal to the angle FBA, for each
of them is a right angle (Ax. 11); add to each the angle ABC.
Therefore the whole angle DBA is equal to the whole angle
FBC (Ax. 2).
And because the two sides AB, BD are equal to the two
sides FB, BC, each to each (Def. 30), and the angle DBA
equal to the angle FBC ;
Therefore the base AD is equal to the base FC, and the
triangle ABD to the triangle FBC (I. 4).
Now the parallelogram BL is double of the triangle ABD,
because they are on the same base BD, and between the
same parallels BD, AL (I. 41).
And the square GB is double of the triangle FBC, because
they are on the same base FB, and between the same parallels
FB, GO (L 41).
But the doubles of equals are equal (Ax. 6), therefore the
parallelogram BL is equal to the square GB.
Li the same manner, by joining AE, BK, it can be shown
th.a.t the parallelogram CL ia eqvial to the square HC.
I'RoPoamoKd. 137
Therefore the whole square BDEC is equal to the two
squares GB, HC (Ax. 2) ;
And the square BDEC is described on the sti-aight line LA^+Aca
BC, and the squares GB, HC upon BA, AC.
Therefore the square described upon the side BC is equal
to the squares described upon the sides BA, AC.
Therefore, in any right-angled triangle, &c. Q. E, D.
Proposition 48.— Theorem.
If the square described upon one of tlie sides of a triangle
he equal to tits squares described upon t/te otiier two sides if
it, the angle contained by tJiese two sides is a right angle.
Let the square described upon BC, one of the sides of the
triangle ABC, be equal to the squares describeil uj)on the
other sides BA, AC;
The angle BAC shall be a right angle. Draw
Construction. — From the point A draw AD at right right
angles to AC (I. 1 1). ''g*^ ^
Make AD equal to BA (I. 3), and join DC. (Do not
Pboof. — Because DA is equal to AB, the square on DA 5!2fj^*
is equal to the square on BA.
To each of these add the square on AC.
Therefore the squares on DA, AC are
equal to the squares on BA, AC (Ax. 2).
But because the angle DAC is a right
angle (Const), the square on DC is equal to
the squares on DA, AC (I. 47),
And the square on BC is equal to the
squares on BA, AC (Hyp.) ; ^ ^^^
Therefore the square on DC is equal to the square on BC dc-j =
^AX. l) \ and
And therefore the side DC is equal to the side BC. DC = BO.
And because the side DA is equal to AB (Const.), and AC
common to the two triangles DAC, BAC, the two sides
DA, AC are equal to the two sides BA, AC, each to
each.
And the base DC has been proved equal to the base BC j
1^8 GEOMETRif.
Hwioe ^ Therefore the angle DAC is equal to BAG (I. 8).
zBAC. But DAC is a light angle (Const.) ^
Therefore also BAC is a right angle (Ax. 1).
Therefore, if the square, &c. Q, E, D,
EXERCISES.
Prop. 1 — 15.
1. From the greater of two given straight lines to cut off a portion
which is three times as long as the less.
2. The line bisecting the vertical angle of an isosceles triangle also
bisects the base.
3. Prove Euc. I. 5, by the method of mper-posUion,
4. In the figure to Euc. I. 5, show that the line joining A with
the point of intersection of BGr and FC, makes equal angles with
AB and AC.
6. ABC is an isosceles triangle, whose base is BC, and AB is
perpendicular to BC ; every point in AD is equally distant from B
andC.
6. Show that the sum of the suin and differehce of two given
straight lines is twice the greater, and that the difference of the sum
and difference is twice the less.
7. Prove the same property with itegard to angles.
8. Make an angle which shall bd three-fourths of a right angle.
9. If, with the extremities of a given line as centres, circles be
drawn intersecting in two points, th^ line joiilins the points of in-
tersection will be perpenoicular to the given Ime, and will also
bisect it.
10. Find a poini which is at a giveii distance Irom a given point
and from A given line.
11. Show that the Sum ol the angles round a given point are
together equal to four right angles.
12. If the exterior angle of a triangle and its adjacent interior
angle be bisected, the bisecting lines wul be at right angles.
13. It three pamts, A, B, C, be taken not in the same stnli^ht line,
and AB and AC be joined and bisected hy perpendiculars ivhidi meet
in D, show that DA, DB, DC are equal to feacn other.
Prop. 16—32.
14. The perpendiculars from the angular points upon the opposite
sides of a triangle meet in a point.
15. To construct an isosceles triangle on a given base, the sides
being each of them double the gv\eiib»ae.
feXEBCI^ES ON THE PROPOSITIONS. 139
16. Describe an iso&celes triangle having a given base, and whose
Vertical angle is half a right angle.
17. AB is a straight line^ C and D are points on the same side of
it ; find a point E in AB such that the sum of CE and ED shall be a
minlmnm.
18. Having ^ven two sides of a triangle and an angle, construct
the triangle. Examine the cases when there will be (l.Jone solution;
(2.) two solutions; (3.) none.
19. Given an angle of a triangle and the sum and difference of the
two aides including the angle, to construct the triangle.
20. Show that each of the angles of an equilateral triangle is two-
tfairds of a right angle, and hence show how to trisect a right angle.
21. If two angles of a triangle be bisected by lines drawn from the
mpnlar points to a given point within, then the line bisecting the
third angle will pass through the same point.
22. The difference of any two sides ot a triangle is less than the
tiiirdside.
23. If the angles at the base of a right-angled isosceles triangle be
Uaected, the bisecting line includes an angle which is three halves
of a right angle.
2L The sum of the lines drawn from ally point within a polygon
to the angular points is greater than half the sum of the sides of the
polygon.
1>R0P. 33—48*
25. Shovt that the diagonals of a square bisect each other at right
•n^es, and that the square described upon a semi-diagonal is half
the given square.
26. Divide a given line into any number of eyial parts, and hence
show how td divide a line similarly to a given line.
27. If D and E be respectively the middle points of the sides BC
and AG of the triangle ABC, and AD and BE be joined, and intersect
in 6, show that GD and GE are respectively one-third of AD and BE.
28. The lines drawn to the bisections of the sides of a triangle
from the opposite angles meet in a point.
29. Describe a square Which is live times a given sqtiare.
30. Show that a square, hexagon, and dodecagon Will fill up the
ipace round a point.
31. Divide a square into three equal areas, by lines drawn parallel
to one of ihe diagonals.
32. Upon a given straight line constnict a regular octagon.
33. Divide a given triangle into equal triangles by lines drawn from
one of the angles.
3i. If any two angles of a quadrilateral are together e^ual to two
right angles, show that the sum of the other two is two n^ht angles.
35. The area of a trapezium having two parallel sides is equal to
half the rectangle contained by the perpendicular distance between
the parallel sides of the trapezium, and the sum of tho paT9bile\ ^V^^'^.
140 GEOMETBT.
36. The area of any trapezium is half the rectangle contained by
one of the diagonals of the trapezimn, and the emm of thepeix>endica-
lars let fall upon it from the opposite angles.
37. If the middle points of the sides of a triangle be joined, the
lines form a triangle whose area is one-fourth that of the given
triangle.
38. If the sides of a triangle be such that they are respectively the
sum of two given lines, the difference of the same two lines, and
twice the side of a square equal to the rectangle contained by these
lines, the triangle shall be right-angled, having the right angle
opposite to the hrst-named side.
39. If a point be taken within a triangle such that the lengths of
the perpendiculars upon the sides are equal, show that the area of the
rectangle Contained by one of the nerpendieulars and the perimeter of
the triangle is double the area of tiie triangle.
40. In the last problem, if O be the given point, and OD, OE, OF
the respective perpendiculars upon the sides BO, AC, and AB,
show that the sum of the squares upon AD, OB, and BO exceeds the
sum of the squares upon AF, BD, and OD by three times the square
upon either of the perpendiculars.
41. Having given the lengths of the segments AF, BD, OE, in
Problem 40, construct the triangle.
42. Draw a line, the square upon which shall be seven times the
square upon a given line.
43. Draw a line, the square upon which shall be equal to the sum
or difference of two given squares.
44. Keduce a given polygon to an equivalent triangle.
^ 45. Divide a triangle into equal areas by drawing a line from a
given point in a side.
46. Do the same with a jgiven parallelogram.
47. If in the fig., Euc. iT 47, tne square on the hypothennse be on
the other side, show how the other two squares may be made to cover
exactiy the square on the hy]>othenuse.
48. The area of a quadrilateral whose diagonals are at right angles
is half tiie rectangle contained by the diagonals.
SECTION III.
ALGEBRA.
CHAPTER I.
SLEMENTABT PRINCIPLES.
L Algebra treats of numbers, tbe numbers being repre-
sented hy letters (symbols of qtumtity)^ affected with certain
symbols of quality ^ and connected by symbols of operation.
It is easy to see that these symbols of quantity may be
dealt with very much as we deal with concrete quantities in
arithmetic. Thus, allowing the letter a to stand for the
number of tmits in any quantity, and allowing also 2 a, 3 a,
4 a, &C., to stand respectively for twice, thrice, four times,
tc, as large a quantity as the letter a, it at once follows
that we may peiform the operations of addition, subtraction,
multiplicatijon, and division upon these symbols exactly as we
do in ordinary arithmetic upon concrete quantities. For
instance, 4 a and 6 a make 10 a, 9 a exceeds 5 a by 4 a, 15 ce
18 5 times 3 a, and 7 a is contained 8 times in 56 a.
Neither is it necessary in these operations to state, or even
to hfiow the exact number of imits for which any symbol of
quantity stands, nor indeed the nature of these units ; it is
simply sufficient that it is a symbol of quantity. Thus, in
the science of chemistry, we use a weight called a critii ;
and a person unacquainted with chemistry might not know
whether a crith were a measure of length, weight, or
capacity, or indeed whether it were a measure at all, yet he
would at once allow that 6 criths and 6 critba are ill criths^
that twice 4 criths ti,v^ 3 criths^ <t(^,
U2 AL6EBEA.
The Signs + and - as Symbols of Operation.
2. In purely arithmetical operations, the signs + and -
are respectively the signs of addition and subtraction. _In
this sense, too, they are used in algebra. i
Thus, a + b means that 6 is to be added to a, and a ^T>
means that 6 is to be 8t4htracted from a.
Hence, as long as a and h represent ordinary arithmetical
numbers, a + b admits of easy interpretation, as also does
a ^ b, when b is not greater than a. But when b is greater
than a, the expression a - & has no arithmetical meaning.
By an extension, howevw, of the use of the signs + and
•^,we are able to give such expressions an intelligible signi->
fication, whatever may be the quantities represented by
a and 5,
Positive and Negative Qnantlties.— The Signs -h and ^
as Symbols of Affection or Quality.
S. Def. — ^A positive quantity is one which is affected
with a + sign, and a negative quantity is one which is
affected with a - sign.
Let BA be a straight line, and O a point in the line;
and suppose a person, starting .
from O, to walk a miles in the ° "*•
direction OA. Suppose also another person, starting from
the same or any other point in BA to walk a miles in the
direction OB. These persons will thus walk a miles each in
exactly opposite directions. Kow, we call one of these direc-
tions positive (it matters not which) and the other negative.
Let us take the direction OA as positive. We then have
the first person walking a miles in a positive direction, and
the second walking a miles in a negative direction. We
represent these distances algebraically by + a and — a
respectively.
It will therefore be seen that the signs + and — have no
effect upon the magnitudes of quantities, but that they express
the quoMty or affection of the quantities before which they
stf^nd.
THE SUM OF ALOEBBAICAIi QUANTITIES. 143
Again, suppose a person in business to get a profit of £6,
wliile another suffers a loss of £6. We may express these
facts algebraically in two ways. We may consider gain as
positive, and has as negative gain, and say that the former
has gained + 6 pounds, while the latter has gained -^ 6
pounds. Or we may consider loss as positive, and gain as
negative loss, and say that the former has lost — 6 pounds,
while the latter has lost + 6 pounds. We hence see that
the gain of + 6 is equal to a loss of - 6, and that a gain of
— 6 is equal to a loss of + 6.
The Sum of Algebraical Quantities.
4. Let a distance AB be measured to the rigM along the
line AX. And let a further — j s ir --r
distance BC be measured from ^ ^ ''
B in the same direction. By the svm of these lines we mean
the resultiDg distance of the point C from the original point
A, that is to say, the distance AC.
(It may be remarked that we add the line BC to the line AB by
measuring BC in its ovon proper direction from the extremity B of
AB. It is hardly necessary to remind the student that both lines
are in the same straight line AX. )
Let us represent the distances AB and BC by + a and
+ h respectively ; then the algebraical sum of the lines will
be represented writing these quantities side by side, each
"with its ovm proper sign of affection.
Thus the sum of the distances AB and BC is expressed by
+ a + &, or, as it is usual to omit the + sign of a positive
quantity when the quantity stands alone or at the head of
an algebraical expression, the sum of AB and BC is expressed
by a + ft.
Hence, the interpretation of a + ft is that it represents the
distance AC.
Again, taking as above + a to represent the distance AB
along the straight line AX, and \ v. r
measured to the rights let adis- ^^. ^
tanoe BC be measured from B in ' ^ *
the same straight line AX, but — ^- — • ■•■• —
fhis time to the leftf
» I . '
144 ALGEBRA.
Let the latter distance be represented by - b.
Tben, on the principle above, AC is the sum of these dis-
tances, and this sum is represented algebraically by + a ^ b
or a - ft.
(It will be seen that the distance BG is again measured from B in
its ownpro})er direction^ and that the resultant distance AC of the
point C is again the sum of the line AB and BC.)
There will evidently be three cases, viz. : —
1. When the distance BC is less in magnitude than AB,
in which case the point C is on the right of A, and the dis-
tance AC is positive.
2. When the distance BC is eqv^al in magnitude to AB,
in which case the point C coincides with A, and the distance
AC is zero, ^
3. When the distance BC is greater in magnitude than
AB, the point C being then on the left of A, and the distance
is negative.
Now, a — 5 in all these cases represents the distance AC.
It therefore admits of intelligible interpretation whether b
be less than, equal to^ or greater than a.
And, since the distance AC is obtained in the first two
cases by subtracting the distance BC from that of AB, and
in the second case by subtracting as far as AB will allow of
subtraction, and measuring the remainder to be subtracted in
an opposite direction, it follows that —
The sign - , which, standing before a letter, is a symbol
of quality, becomes- at once a symbol of subti^tion in all
cases when the quantity in question is placed immediately
after any other given quantity with its proper sign of
affection.
Hence also we may conclude that the ad4ition of a nega-
tive quantity is equivalent to the subtraction of the correa-
ponding positive quantity.
6. We may prove in a similar way that —
The subtraction of a negative quantity is equivalent to the
addition of the correspondiDg positive quantity.
Let, as before, + a represent the distance AB, measured
from the point A to the right, t
and let it be required to svlh "^ b c
fraci, from f a the distance represented by - 6,
THE SUM OF ALGEBRAICAL QUANTIFIES. 145
Nov, in the last article, we added a distance to a given
distance AB, by measuring the second distance in its aum
directum from the extremity B. We shall therefore be con-
sistent if we atAtract a given distance — 6 by measuring this
distance in a direction exactly/ opposite to its own direction,
from the same extremity B.
Now, the direction of — 6 is to the hfi. If, therefore, we
measure a distance BC to the riglu, equal in magnitude to
the distance to be subtracted, we obtain a distance AC which
h correctly represented by a - { - h). But AC is also
correctly represented by a + b, and hence it follows that
a-(-6) = a + 5.
We may apply the above principle to all magnitudes which
admit of continuous and indefinite extension; as, for instance,
to forces which pull and push, attract and repel ; to time
past and time to come, to temperatures above zero and below
zero, to money due and money owed, to distance up and dis-
tance down, ^., in all which cases, having represented one
by a quantity affected with a + sign, we may represent the
odier by a quantity affected with a - sign.
6. In expressing the sum of a number of qiumtities, the
order of the terms is immaterial.
We will take, as our illustration, a body subject to various
alterations of temperature, and we will suppose the tempera-
tare of the body, before the changes in question, to be zero
or 0^. Let the temperature now undergo the following
dumges — ^viz., a rise of a°, a fall of 6°, a fall of c^ and, lastly,
a rise of (T. Let us consider a rise as positive, and therefore
a fall as negative. We may then represent these changes
leflpectively by + a, — ft, — c, + d.
And it is further evident that the resulting temperature
will be represented by the sum of these quantities, which, as
previously written, will bea-6 — c + <f.
But again, it is plain that the resulting temperature of the
body will not be affected if these changes of temperature take
place in the reverse order, or in any other order. Thus,
suppose the temperature first falls c^, then rises a^, then rises
(T, and, lastly, falls 6^ it is evident that the final temperature
will be the same as before. And the sum of the quantities
^Cf + a, + d, — t, represents t¥9 fiiWil tem]jeratiu;i;^,
6 ^
14G ALGEBRA.
Now, expressing the sum of these quantities by writing them
(Art. 4) side by side with their proper signs of affection,
and in the order in which they stand, we have -^ c + a + c^
— b for the sum.
It therefore follows, since we might have chosen any other
order of these terms with a similar result, that the sum of
any number of quantities, + a, — ' ^, — c, + d, may be
expressed by writing the terms side by side with their proper
signs of affection in amy order whatever.
Nevertheless, for convenience, and for other reasons, we
write the terms generally in cdphaheticaZ order , or we arrange
them according to the power (see Art. 16} of some particular
letter.
7. We may sum up the results and remarks of the last
four articles as follows : —
1. Positive and Tiegative are used in exactly opposite
senses.
2. The sign + before an algebraical quantity affirms the
quality of the quantity as represented without the sign in
question.
Thus, + ( + a) = + a, and + (— 6) = — 6.
3. The sign — before an algebraical quantity reverses the
quality of the quantity as represented without the sign in
question.
Thus, — ( + a) = — a, and — ( - 6) = +6.
4. The algebraical subtraction of a quantity is the same as
the addition of the quantity with the sign of affection
reversed.
5. The algebraical «*addition of quantities is expressed by
writing the quantities down side by side with their proper
signs of affection. And they may be written down in any
order, though we generally write tiiem in alphabetical order,
or arranged according to the power (Art. 16) of some
letter.
Brackets,
8. Brackets — ( ), { (, [ ] — are used, for the most part^
whenever we wish to consider an algel3raical expression con-
taining more than one term as a whole.
BRACKETS.
m
Thus, if \7e wish to express that thd qTiantity 3 a + 7 6 is
to be added as a whole to 4 a, we write-^-
4a + (3a + 7 6),
and, while inclosed within brackets, we think and speak of
3 a -^ 7 6 as OTie quantity^
Again, if we wish to express that & — o is to be subtracted
firom a^ we write—
a — (5 - c).
liot us consider what is Ihe result of subtracting (5 — c)
from Ob We may evidently, if we please, subtract h firs^
then afterwards - c from the quantity so obtained, without
affecting this result
Now, we know by Art. 7 (4.) that this is equivalent to
adding the quantities * h and + c successively.
Now, the sum of a, — 6, + c, is a - 6 + c.
We have therefore a — (6 - c) = a - 6 + c.
We observe that the sign of h within the brackets is -r,
and that of c is «> , whereas, in our final result, these signu
are both reversed. And we hence arrive at the following
important principle ; —
When a minus sign stands before a bracket, its effect on
removing the brackets is to reverse the sign of affection of
every term within.
And it is evident that we may, by a similar course
of reasoning, arrive at a principle equaUy important,
viz. : —
When a plus sign stands before a bracket, its effect on
removing the bracket is to affirm the sign of affection of
every term within.
We shall show, in Art. 9, the use of brackets in expressing
the product or quotient of quantities.
IJiough, as stated above, brackets are, for the most part,
used to group together a whole a nimiber of quantities,
they are sometimes used to inclose single terms. Thus, in
Art.' 6, we have the expression a — (- b). Now, the
brackets are used here to express that the negative quantity
is to be subtracted as a negative qiiantity. And, in liie same
way, the expression a + (—6) indicates that the negative
qwmUity ( - 6) is to be added to the quantity a. When one
jisir only is required we generally use the brackets ^^ \ \i^
148 ALGEBRA.
however, a quantity already in brackets is to be inclosed in
a second pair^ we use { (, as in the expression —
3a- |66+(4c-d)|.
If a third pair be required we use the brackets [ ], and
finally, we sometimes find it convenient to group a number
of terms by means of a vincvlumy thus —
4a; - \^x - {Sy + (3« - 7a; - y)} + 2yJ.
It must be remembered that the wiwalum has in an
expression exactly the same force as brackets.
9. We shall, in this Article, show how to find the value of
a few algebraical expressions, as illustrations of the foregoing
principles : —
Ex. 1.— If a = 1, 6 = 2, c = 4, find the value of
3a + 56 + 7c.
We have only to substitute the value of the letters in the
given expression, putting a sign of multiplication to avoid
ambiguity.
Thus we have —
3a + 56 + 7c = 3xl + 6x2 + 7x4.
= 3 + 10 + 28 = 41.
Ex. 2.— If a; = 5, y = 2, « = 6, find the value cf
4a; + 3y - 9«.
Here, 4a; + 3y-9« = 4x6 + 3x2-9x6
= 20 + 6-54
= 26-54.
The negative quantity is here the larger, and exceeds the
positive quantity by 28, and hence (Art 4 (3),) the result will
be negative.
We therefore have —
4a; + 3y-9»=-28.
Ex. 3.— If a; = 1, y = 3, « = 0, find the value of—
3a; - |2y + (6« -5a; - y)|.
The given expression —
= 3xl-|2x3+(6x0- 5x1-3) |
-3-} Q + io-JT^).^
PRODUCT OP TWO OR MORE QUANTITIES. 149
= 3 - j 6 + (0 - 2) I .
= 3- (6 -2) = 3-4= -1.
It may be advisable to simplify expressions of this kind
before substituting the given value of the letters. The
method of doing tiiis, however, is deferred till we come to
Chapter II.
Ex. I.
K a = 1, 6 = 2, c = 3, c^ = 0, e = 4, find the value of the
following : —
1. 4a+2 6, 36 + 7c, 6a + 4c?, 4c — 7e.
2. a + b + c, a — h + c, b + c-a, a + b — c.
3. 3a + 75 - 4c, 2a + 7d + 3c, 7a - 106 + 2 c.
4. 6a- (2 6- 3c), 76 + (2a - ^d).
5. 3c + (6a - 76 + c), 26 - | 7c + (4c^ - 6) | .
6. |76 - (2c^-c)| - (46 - c + e).
If 05 = 2, y = 3, » = 4, find the sum and difference of the
following expressions : —
7. 3 a; — 4 y and 3 y - 4 x.
8. 7 (a; - y) and 4 (y - z),
9. 3x - {7 y + 4:z) and 8y + 5« — 3aj.
10. 12y - 6y + 4« and 12y + 6y + 4«.
11. X - y - z and x - (y - z).
12. X - ( - 3 y) and y - | 3 a; - ( - 3z)l,
Product of Two or more Quantities.
10. The product of two or more quantities may be
expressed in several ways.
Thus, the product of a, 6, c may be written as follows : —
1. a6c, by placing the letters side by side without any
tifjn between them.
2. a X 6 X c, by i)lacing between them the ».gii x ,
3. a . 6 . ^ hy placing a, dot between them.
150 Algebra.
4. (a) (h) (c), by inclosing the quantities in brackets,
writing them without any sign between the brackets.
When, however, the quantities are negative, or either of
them consists of more than one term, it is best to inclose such
quantities in brackets, and, in most cases, it is necessary to
do so.
Thus, the product of a, - 5", c would not be correctly
expressed by a - 5c (for this means, that the product of
b and o is to be subtracted from a), but must be written
a (- 6) c. Again, the product of 2 a + 3 6 and a + 6b
cannot be written 2a + 3ba + 5 5, as this expression means
that three times the product of 5 x a is to be added to 2 a,
and then 5 6 to the result. The product is correctly ex-
pressed thus —
(2 a + 3 6) (a + 5 b).
(The fitudent cannot be too strongly cautioned agamst leaving out
brackets in cases of this kind.)
The Order of the Letters.
11. tt is evident that a times b = b times a ; for, if we arrange
a rows of b things so as to have a horizontal rows and b
vertical columns, we may either consider the number of rows
or the number of columns. In the former case we have
a times b things, and, in the latter, b times a things.
We may therefore write the letters whose product we
wish to express in any order.
Thus, the product of a, 6, c may be written in either
of the following ways : —
abc, acb, baCj cba, bca, coh.
For convenience, however, and for reasons which the
student will see as he proceeds with the subject, we write
them in the order of the alphabet, tmless there happen to be
special reasons to the contrary.
*
Rule of Signs in Multiplication.
12, It was shown (Art. 3^ that a minus sign does not affect
the jQEiagiiitude of a quantity, Wt ^vm^'^ \fe8» ^^ectixm or
QUOTIENT? OB* TWO QUANTITIES. 15l
tjfuallty* It therefore follows that the product of + a and
- h will be the same in magnitude as that of + a and + hy
or of a X b — ^that is, will be equal in magnitude to ah.
But it ia evident that the quality of the product will be the
same as that of - b, for it is a times a negative quantity.
We therefore have + a ( - 6) = ^ ab.
So also, the product of - a and + b, will have the same
magnitude as that of + a and + b — that is, its magnitude
will be ah. But since, to take a geometrical illustration, a
line of length, 6, drawn negatively once, negatively twice, &c.,
must give a negative result, the quality of the product in
question must be negative.
"We therefore have ( - a) ( + 6) = - a6.
Again, the product of - <x and - b will be equal in
magnitude to oh, and its quality will be evidently the
reverse of the quality of the product of - a and + b, and
the quality of the latter product is above shown to be
negative. The product of -ax - b will be therefore
positive.
Hence we have ( - a) ( - 5) = + ab.
Collecting these results, and remembering that (+ a)
^ + 5) = + a6, we have —
( + a) ( + 6) = + ah.
(+ aW- 6) = - ab.
( - aW + 5) = - ab.
( - a) ( - 6) = + ab.
We have then the following rule of signs in multiplica-
tion : —
Rule. — ^Like signs give + , and unlike signs give - .
Qaotient of Two Quantities.
13. The quotient of two quantities is expressed in either
of the two following ways : —
1. By placing the divisor under the dividend, separated by
a line.
Thus, the quotient oiahyb = i^.
b
15^ AL(iifiRAi
2. By placing the divisor after the dividend with a sign
-f between them ; thus, a -r- b.
When either of the quantities is negative, it is better, if the
second be used, to inclose the negative quantity, if the
divisor, in brackets.
Thus, the quotient of a by - 5 may be written a -f - 6,
but it is better written a -f (- h).
And when either of the quantities contains more than one
term, it must^always be inclosed in brackets when expressed
by the second method.
Thus, the quotient of a + 2 6 by 2 c = (a + 2 6) -f 2 c,
not a + 2 6 -f 2 c, for this would mean that the quotient of
2 5 by 2 c is to be added to a.
Rule of Signs in Division.
It is evident from the last article that —
( + a6) -f ( + a) = + h.
(-a5)
if:
a) = - b.
(- ab) -T- {- a) = + b.
1+ ab) -T {- a) = - 6.
*We have therefore the following rule : —
Rule. — ^In division, as in multiplication, like signs give
+ , and unlike signs give - .
CoefSicients.
14. When a quantity can be broken up into two factors,
each of those factors is called a coefficient of the other.
Thus, taking the "quantity 4 abc, we see that 4 is the
coefficient of abc, 4 b that of ac, 4 c that of a6, 4 be that of
a, &c. We call 4 a numerical coefficient; but when the
coefficient contains a letter or letters we call it a literal
coefficient. We often speak of the numerical coefficient of
a quantity as the coefficient of the quantity. It is, moreover,
unusual to write down unity as a numerical coefficient, and
8o, when an algebraical quantity has no expressed numerical
coef^cient, we may, couveraeVy, coi^^w Mmty aa such.
POWERS. 153
16. lAke qiumtities are generally defined as those which
differ only in their rmmerical coefficients. Thus, 9 a;, 3 y,
4 asyz, 12 ah are respectively like to the quantities 10 a;, 5 y,
6 xyz, 4 abf while 3 a?, 6 xy, 7 U* are called unlike.
It is sometimes convenient, however, to consider as like
qucmtities those which may contain perhaps one common
letter only, though containing other letters which are not
common. Thus, for the purposes of addition and subtrac-
tion, we may consider 3 ax and 4 &a; as like qimntities^
having 3 a and 4 b respectively as their coefficients. Now,
their sum, since they are respectively equivalent to 3 a and
4 b times x, will be equivalent to (3 a + 4 6) times x, and
may then be written (3 a + 4 6) a;.
Hence we learn that addition and subtraction of quanti-
ties having any common factor may, considering the un-
common factor as coefficient, be expressed according to the
foDowing rule : —
Add together the coefficients of the commxm factor, and take
the sum as a new coefficient of the common factor.
And a similar rule will apply to the subtraction of such
quantities.
Thus, we have the sum of 4 axy, - 2 bc7/, 3 c^ey,
= (4 aa; - 2bc + 3 de) y.
Powers.
16. We have seen (Art. 10) that dbc means that a is to be
multiplied by 6, and the product by c. And so, if we wish
to express that a is to be multiplied by a, and this product
again by a, we might write the expression thus : aaa,.
Instead of this, however, we usually place a small figure at
the top and on the right hand of the letter a, to indicate how
many times the letter appears as a factor. In this case,
therefore, we write a^ We call quantities of this form
powers of the letter in question; and so, remembering that a
may be considered as a^ on this principle, we have : —
a, the first power of a ; a^y the second power of a ; a^, the
third power of a, &c. The figure written at the right hand
at the top of the letter is called the index or expou^Ut^«kXi"i\\»
is usual to caU a^, a^ respectively the square and c^^]be oi a,
154 ALGEBRA.
from the fact they express respectively the area of a square
whose side is a, and the volume of cube whose edge is a,
17. The square root of a quantity is that quantity which,
when raised to the second power or aqua/redy will give the
original quantity. _
It is generally written ^, Thus, \/l6 = 4, V144: = 12.
The cube root of a quantity is that quantity which,
when raised to the third power or cubed, will give the
original quantity.
It is generally written ^. Thus, aJ^ = 2, ^7 = 3,
Jl7'2S = 12. And so the fourth, fifth, &c., roots are
indicated by the symbols ^^ , l/^ &c., respectively.
18. The dimensions of an algebraical quantity are the
sum of the indices or exponents of the literal factors.
Thus, the dimensions of Za^b^c^ = 2 + 3 + 4 = 9, and
the dimensions of ahaP =1 + 1+2 = 4.
19. A homogeneous expression ia one in which the dimen-
sions of every term are the same.
Thus, a^ + 3 a*6 + 3ab^ + 5' is homogeneous,
Whereas, 5 a + 3 a6 + 7 a^bc is not homogeneous.
Ex. 11.
Ifa = 2, 6 = 3,V= Q,d = \,
Find the values of —
1. 6a^ + 36* - 5c^; a6 + ac + ftc, 6c + 5c? + cd.
2. a' + 3 aJ'b + 3 aft* + i'^ ; a'* + 6* + c^ - 3 abc.
3. (3a + 7 6) {ia - 9 6) ; (a* + W) {a + &)(«-. b).
4. 6 {2 a' - 4 (2 b^ - 2c« - (f)} ; a'b + ab"" + a?c +
ac^ + b^c + b<?.
If a = 1, 6 = ^ 2,c= - 3, c^ = 0, e = 4,
Find the values of —
5. (a + 6 + c + rf + ef; {a? + 2a6 + 6* - c^) + (a +
5 + c).
a* - cZ* c^ -\- ZcH + Zcd^ -^ d^
6.
a* + a^(i + ocZ* + <^^ fi'* - 3 i^c + 3 6c* - c»
^- -^/ - f °: * " : ■J^rz^TTTn-
ab + ao •\- ho
ADDITION, 165
8. 3(45 + 5cy + 4(c +ey; abcde,
Jfx = 3, y = 4, » = 0,
Find the value of —
9- (3a; - JWVff (2A + ^aP + y" + 2).
10. {5a? + 2(y + 2)^} {5ar»- 2{y + zf}.
11. a:* + y* + «*.
12. (o^^f)^ ^ {Zo^ + 3(3ar» + ^xy ■\- f)y].
CHAPTER 11.
ADDITION SUBTRACTION, MULTIPLICATION, AND DIVISION.
Addition.
20. EuLE. — Arrange the terms of the given quantities so
that like quantities may be under each other ; add separately
the positive and negative coefficients of each column ; take
the difference and prefix the sign of the greater, and annex
the common letter.
^ (When the coefficients are all positive or all negative, we, of course,
simply add them together and prefix the common sign for the coeffi-
cient of the sum.)
Ex. 1. Ex. 2.
3a + 56-3c -3a + 76+ c-4c^
2a-76 + 4c 2a-.25 + 5c+3c?
5a+ 5-2c -7a-.36 + 2c-6c^
4a-36 + 8c 2a + 66-8c+6dJ
Ans. 14a-46 + 7c Ans. -6a+76 - d
Ex. 3. Add together 5ar*-32/^ + 3y, 62/* + 7a^-4a;,
4a!y + 6a;-5, -2a:;^-3ajy+2.
Arranging like quantities in each expression under each
other, we have : —
5ar» - 32/* + ^y
7xy + 6y^-'4:X
4xy + Qx - 5
-2a? - 3xy A- 1
Ans, 3a^ + 8x^ + 3 y^ + 2 x -V-3^ — ^ \.
156 AtiQEfi&A.
Ex. III.
Add together- ♦
1. 3a - 2b, 4: a + 7 h, 2 a + 3h, a - 5 b,
2. da" + 1b\ - Sa^ + 45^ a« + b\ 4:d? - 1261
3. a + 6 + c, 3a + 26 + 3c, -4a+76-c, 26 + 5c.
^, X — y-z, y — x-ZyZ-x-y^x + y + z, ^
6. 3a^ - 4a6 + 6 6^ 7a5 - a^ - 6^ 2a2 - 3a6 - 4 b%
ia^ + ah -b\
6. 2aJ* - la? + 3, - 4a^ + Gic^ _ 2a; + 7, jc^ - 2 a:» - 4 a;,
Gaj'-Qa; - 12.
7. 2a2 + 7a6 + 362- 6a~66 - 2, a2+ 3a- 26 + 9,
9a6 -2a-36 + 4, -3a2-.i2a6 - 3 62+5a+ 106- 15.
8. as* - a?y - a?z + xy^ - xyz + xz^, o^y - a^ - xyz +
^ — y^z + 2/?, a:^« - ocyz - ocz^ + y^z - y;r + ;?;*.
9. aJ* + Q^y^ + as'y, - as^y — o?y^ - a^, y* + oc^ + a:^^.
10. a' + a62 + oc^ + 2 a26 - 2 a^c - 2 a6c, c?b + 6' + 60^ +
2a6* - 2a6c - 26^0, a^c + 6^0 + c* + 2a6c - 2a^ - 26c*.
11. 0^ " xi^ + XS? - 3a?y + 3ar'«, 3ar^3/» + 3ar';s2 + 3a^2J
- 3 xyz^ - 6 oi?yz, y* - a?y - y:^ + 3 o?y^ - 3 o^yz, - 3 a;2^
- 3 xyT? — 3 y« + 3y«2 + 6 a;^^*^, ;2i* + i?;s - y;s - 3 ai^y;^ +
Z:^:^, 3a^« + 3ax2;» + 3^^- 3yz^ - 6xy:??.
12. a* - a'6 + 3 aV 4- a62c - 3 060* - 6»c, a»6 - a'c +
3 060* - 3a<? - b^'c' + b% a«c - a'c? + 3 ac* - 3 ac'd + 6V
- 6W, - a* + a'c^ - 3 aV + 3ac'd - ab^c + 6W.
Subtraction.
21. We have seen, Art. 7 (4.), that the subtraction of a quan-
tity is equivalent to the addition of the same quantity with
its sign of affection reversed. We therefore have the follow-
ing rule :—
Rule. — Change the sign of each term of. the subtrahend,
and proceed as in addition.
Ex. 1. Ex. 2.
bx + Qty - 3z 6a* - 3a6 + 46'
2a! " 2y + 2z - 2c? - 3ah - 26*
BRACKETS. 157
Ex. 3.
2 ar* + ^xy^ + 3y^
a^4-3ar'y4-3r^+2i /'
ar» - 3 a^y + 3 ic/ + ^.
Ex. lY.
1. From 6a + 76 + 3c take 2a + 56-2c.
2. From 2x — ^y - ^z take Qx-by — 2z,
3. Take5a2 + 3a6 + 462 + 3a+76 + 8from6a' + 3J'-2a.
4. Take 6a* + ^aV + aJ*from 8a* + 6aV + 2ic*.
5. Subtract the sum of the quantities a* + 2 a^6^ + 6*,
a^^2a^h^ + 6*from6a* + 8a'6' + 6 6*.
6. From a^ + y^ + «^ — 3 a3y« take 4:tc' + 2/'+ 4:«*+ 3a3^;2;
+ 30^^ - 3icy5?.
7. From3aJ* + ?aa'- 9aV + a'a;-a* take 2a;* + ^ax"^
+ 4 a'a; + a\
8. Take a» - 5 a^ft + 7 a6* - 2 5' from the sum of 2 a' -
9a^b + lla62 - 368 and 6'- 4a62 + 4a^6 - a'.
9. Subtract a + h-}-c + d from e +/ + ^^ + A.
10. Take aJ* - 4 a^y + 6 ar^ - 4 a^ + y* from a;* + 4 ar'y
+ 6 a^j^ + 4 xi^ + ^, an'l subtract the result from their sum.
11. Add together the given quantities in the last example,
and subtract the result from 3 sc* + 10 aPy^ + 32/*.
12. Take a^ -h b^ + <? ■{• 2 ab + 2 ac + 2bc from 2 a' + 2 6'
+ 4 a6 - cr*.
Brackets — continued.
22. It was shown, in Art. 8, that, when a quantity inclosed
in brackets is to be added, we may remove the sign ( + ) of
addition and the brackets without changing iihe sign of the
terms within the brackets. On the contrary, when the
quantity in brackets is to be subtracted, or has the sign
minus before it, we must change the sign of every term
within the brackets on removing the brackets and iiie sign
of subtraction.
We shall now see how to simplify expressions involving
brackets ogjmeoted Jb^ fib^ ai^ of a4^tion aud^u\i\?t«i^v«x\--^
158 ALGEBRA.
Ex. 1. Simplify (3a+5 6)-(6 6-2c) + (- 2a + 6
- 3 c).
The given expression = 3fl& + 56-66 + 2c-2a+6-
3 c, or adding together the like quantities,
= a - c,
Ex. 2. Reduce to its simplest form —
a - (6 - c) + I 6 + (a - c) I - | (a - 5) - c | ,
(When a pair of brackets is inclosed withia another pair, it is con*
venient to remove the inner one first.)
Hence the given expression —
= a-6 + c+6 + rt-c-a+6 + c = a + 6 + c.
*, « «. ,./..! . 3 a; X - 7 6 a; - 9
Ex. 3. Simplify the expression --: — + — - — .
The line separating the numerator and denominator of a
fraction is a species of vinculum, since it serves to show that
the whole numerator is to be divided by the whole denominator.
Hence, on breaking up the two latter fractions into fractio&s
having one term only in the numerator, we have —
Sx a;-? 6a;-9 3 17 6 9 19
— — J- =-a3— - X + - +— X"— = x+ —
4 2 8 4 2 2 8 8 8.
23. As it is often necessary to inclose quantities within
brackets, we shall now show how this is done.
The following rule needs no explanation : —
E.ULE. — ^When a number of terms is inclosed within
bracTcets, if the sign placed before the brackets be + , the
terms must be written down with their signs of affection
unclumged ; but, if the sign placed before the brackets be - ,
the sign of affection of every term placed within the brackets
must be chcmged.
Thus we may express a + h - c - dia any of the follow-
ing ways : —
a + 6-c-(f = a+ (6-c-cf) = (a+6) - (c + d)
/ When the word »ign is used in fature^ the student is to under-
tiand sign ojaffection^ unleBa otUerwiaQ expressed.)
MULTIPLICATIOir. 159
Ex. Y.
Simplify the expressions —
I. 3 X - {2 X - y) + {5x + 3 y -^ 2 z) ^ (7 X - 4: ij ^ 3 z).
3. (2 a\ + 2 a-6 + 2 a^) - (3 a' + arb + ah^ - b^) + («»
3. 1 -- (1 - 2 ic) - j 3 - (4 ^ 3 cc) j + j 5 - (4 a: -
4. 6ar»-(2-3aj + fB2)+ | -7 + (5aj-8a;-2)|
- (3 - 3 - 34
Group together the terms of the four following expressions,
BO that— -
(i) The first two and last two are inclosed in brackets.
(ii) The last three are inclosed in brackets.
(iii.) The first three are so inclosed, and an inner pair of
brackets used to inclose the second and third.
5. a — b + c -^ d.
6. -6a + 76-3c + 5d
7. -- 4:0? + I2ix?y - 12aJ2/» + ^f-
8. a» - Z>3 « c» + 3 abc.
Add together —
9. oai^ + bxy + cf, - c^ar^ - aocy + e'lfy 6ar* + 3 cocy + fy^,
- 2aa? - 2%2.
10. ojx " cy ^ eZy ^ bx -{- dy -{- fz, ex - ey - gz, - dx +
fy + hz.
Subtract —
II. {2 a + b)x - (3 b ■{• c)y + {4:C ^d)z from {3 a - d) x
•h (4:b - a) y + {5 c " b) z.
12. (1/ - z) a^ + (z - x) ab •{■ (x -^ y) b^ from {y - x) c?
'^ {y - z) ab - (z " x)b\
Multiplication.
514. Remembering the definition (Art. 14) of a coefficient,
it follows that the product of two terms having coefficients
is found by multiplying the product of the coefficients by the
product of the remaining factors.
ThuB; the product of 4 a and 3 &, or 4 a x 3 {» = \2iab%
160 ALGEBRA.
Again, the terms to be multiplied may contain like
letters.
Now, a' = aouiy a^ = aaacm, and hence it follows that
0? X a^ = aaa x aaaaa = aaaaaaaa = a^
And, generally, we have
a** X a** = aaa,,. to m factors x aaa... to n factors,
= oaa... to (m + n) factors = a*" "•" **.
It therefore follows that the product of the powers of like
letters is found by adding together the exponents of the like
letters.
Thus, a^ X a^ = a', and a^ x a = a*, &c.
And, further, as regards the sign of the product, we have
seen, in Art. 1 2, that like signs give + , and unlike signs
give -.
There are three things, then, which must be attended to
in the multiplication of algebraical terms, viz. : —
1. The sign>s. — Like signs give +, and unlike signs
give ".
2. The coefficients, — These are to be multiplied like ordin-
ary numbers.
3. The letters, — ^The exponents of like letters are to be
added together, and the powers so obtained written side by
side with the unlike letters.
Ex. 1. Ex. 2. Ex. 3.
6 a^6 - 3 ocyz - 5 ahd
2a^c ^y^ - 2hcd^
Ans. \2a^hG Ans. - ^;^fs^ Ans. lOaJy'cdJ^
25. Whenever the multiplicand or multiplier, or both,
contains more than one term, it is evident the product is
found by multiplying each term of the multiplicand
separately by each term of the multiplier, and adding
together the separate products.
Ex. 4. Ex. 6.
3a2 - lah +562 a? - Za^y + 30:3^ - y^
6 06 " 2xy
Am,lQ<^'{^^i%g?})^\i^i^\ ,- 3(C*y + 6a:»y^-6a:*y»+3a^
MULTIPLICATION. 161
Ex. 6.
a? + ocy + y'^
JB* — ajy + ^
a;* + a^y + a^^
- a»2/ - a^y^ - asy*
Ans. ic* + o^^y* + y*.
Ex. 7.
aj? + (a + 6) a; + o5
X + c
a^ + (a + 6) ar* + o&c
ca^ + (oc + ftc) a; + otc
a:* + (a+6+c)a^ + (a6 + ac + ^)aj + o^.
It will be observed that the terms in the above examples
are all arranged according to the power of some letter.
Thus, in Ex. 4, they are arranged according to the descend-
ing powers of a; and in Exs. 5, 6, 7, they are arranged
according to the descending powers of x. It matters not
whether they are arranged according to ascending or descend-
ing powers, and the result would be the same if they are not
80 arranged. It is then, however, much easier to collect like
terms, as they generally fall under each other. When we
come to division we shall find it necessary to arrange the
terms according to the power of some letter.
Ex. VI.
1. Multiply 3a + 25by4a-3 6, and 6 a; + 7 y by
3 a; — 5y.
2. Multiply a5* + 2a^- 2^ by a;- 2y, and 15 a:;* +
17ay - 43^ by 2aj + y,
3. Multiply a^ + 2 06 + 5' by a' - 2 ab + h\ and o*
+ ft» by a« - h\
4. Multiply a* + a:?y + a^ + y* by a; - y, and the pro-
duct by a^ + ^.
5. Multiply o' + 6^ + 0* — a6-ac-^ by a + 6+c.
6. Rnd the continued product of aJ* + j/*, a? -V -jf , x -v )j^
162 Algebra.
7. Develop the expressions {a + 6)', and {x - y)*.
8. Multiply5a'+ 15a«+ 45a + 135bya»- 2a - 3.
9. Multiply l + 3a?-7a:?bya?-2, and a - xhj a? +
ax -h a\
10. Multiplya* - 2a»6 + Sa'^J' - 2a6' + 6*bya* + 2a^6
+ ZaJ'U' + 2ah^ + h\
11. Multiply a + hx -{- CO? + da? hy ex + /.
12. Find the product o£a? + px + q and x - a,
13. Find the continued product of x + a, x + h, x + c,
26. We have explained in the last Article the general
method to be pursued in Multiplication. There are,
however, many algebraical espressions which may be mul-
tiplied by inspection.
L Expressions of the form (a + 5)'.
It is easily found by long multiplication that
{a + by = a" + 2ab + h\
and this, expressed in ordinary language, may be read thus : —
The aqua/re of the sum of two quantities is the sqaaire of
the first ^ jilus twice their prodiict, plus the sqtiare of the
second,
(This rule is evidently true whether the quantities are poaUive or
negative,)
Ex. 1. (3a + 7bY = (3a)2 + 2 (3a) (76) + {7hy = 9a'
+ 42 a6 + 49 6^.
Ex. 2. (6a; - 6)^ = {Qxf + 2 (6a;) ( - 6) + ( - 5)^ =
ZQa? - 60aj + 25.
Ex. 3. (a + 6 + c)^ = I (a + h)'+ cV
= (a + by + 2la + h)c + (?
= (a^ + 2a5 + 6^) + 2 (ac + be) + <?
= a* + 6* + c* + 2a5 + 2ac + 2bc.
Kemark. — It is a very common mistake with beginners to write
down a' + 6' as the square of (a + 6), thereby leaving out twice
the product of the quantities a and b. They should impress this for-
mula, viz. : —
{a + 6)« = a* + 2 a6 + 6«
thoroughly upon the nund.
MULTIPLICATION. 163
II. The fonn {a + fe) (a - h).
It may be easily found by long multiplication that
(a + » (a - 6) = a« ^ h\
which, in ordinary language, may be thus expressed : —
The ^product of the sfwm and difference of two qvmUities is
the difference of the aqua/res oftlve qttantities.
This formula may be applied in all cases where the terms
of the quantities to be multiplied are of the same magni-
tude, but when some of them differ in sign.
Ex. 1. (a + 6 + c + cf)(a + 6-c-c?)
= I (a + 6) + (c + cf) I \{a + h) - (c + c?) |
= (a + hf -• {c + df = {a^+2ah + h^) - (cU 2ccZ + dr)
= a* + 6« - c-' - cZ^ + 2a6 - 2cd.
Ex. 2. (a»- 3a=6+ Saft^- Jf) (a'+ Sa^ft + ^ah^+ P)
= |(a» + 3a6^-(3a^6 + 6')| \{a^ + .3a¥) + {Sa'b + ¥)l
= (a» + Zah")^ - (3a^6 + by
= («• + ea^'b^ + Oa^ft*) - (9aW + Qa'b* + 6«)
= af'^Sal'b^ + 3a26*-6«
The principle to be adopted in all such cases is to find
what terms in the given quantities are exactly alike, and put
ikem first in brackets, when the remaining terms will fall
into the proper form. Thus- —
{3a + 7b + 5c - d) {3a-7b + 5c + d) = |(3a + 5c)
+ (7 6 - (f) j j (3 a + 5 c) ^ (7 6 ^ c?) j .
And (3a -76- 5c]+ S) (3a + 7b + 5c + d)= \{3a + d)
- (7b + 5c)j [(3a + rf) + (75 + 5c) j.
(m.) The form (x + a) (x + b).
By multiplication we find that—
{x + a) (x + b) = a? + (a + b) X + abf
which in ordinary language may be thus expressed : —
Theprodtict of two binomiaXa containing a common term and
an uncommon term, is the squa/re of the commmi term, 'plus
the sum of the tmcoTnmon terms multiplied into t]ie ccynvnum
term, plus 1^ produce of the tmcommon term.
164 ALGEBRA.
is
Thus—
(a; + 2) (a; + 3) = ai* + (2 + 3) » + 2 x 3 = a^ + 5 a; + 6.
aj + 6)(aj- 1) = ai=»+(6-l)a; + 6(-l) = ar» + 5x - 6.
(aj-5)(aj-7)=aj='-(5 + 7)aj + (-6)(- 7) =a*-12a;
+ 35.
(aj + 3a) (a; - 5a) = ai* + (3 a - 5 a) a; + (3 a) ( - 5 a) = a?
- 2 oa; - 15 a*.
And so,
(a; + a + 6)(a; + c + cQ = (aj + a + 6)(aj + c + d)
= a? + {a + b + c + d)x+(a + b){c + d).
We may extend this formula to any number of factors.
Thus, by multiplication, we find that —
(a; + a) (a; + 6) (a; + c) = a* + (a + 5 + c) ai* + (ab + ac
+ be) X + abe.
(a; + a) (a? + 6) {x '\- c) (aj + <f) = aJ* + (a + 6 + c + cf)
a? + {ah + a4: + ad+bc + bd + cd)a? + {ahe + oW + acd
+ bed) X + oicc?.
Zaii; of Formation of the Terms,
1. It will be seen that the coefficient of the first term is
in each case unity ; that of the second term, the sum of the
uncommon letters taken singly ; that of the third term, the
sum of the uncommon letters taken two together ; that of the
fourth term, the sum of the uncommon letters taken three
togeHieTy &^,
2. The power of the common letter is in the first term that
of the number of the binomials, and it sinks one every term.
Ex. (a; + 1) (a; + 2) (aj + 3) = aj» + (1 + 2 + 3) a* + (1 +
2 + 1 + 3 + 2 + 3)aj+l + 2 + 3=a»+6ar'+lla: + 6,
(IV.) The form (a + b + c + d + Ac.)'.
By ordinary multiplication or otherwise we have (a + 6 + c
+ c?)' = a' + 6* + c' + cP + 2a5 + 2ac + 2a^+26c +
2bd -\- 2cd; and a similar result follows if we take a larger
number of terms.
We may hence deduce the following rule : —
The square of the sum of any number of quantities is the
sum of the squares of tlie c^Maxit\t\e» together with twice the
sum of the products foim^ \>y m\3li\i^yfia%^^^BK*» o^iantity
MULTIPLICATION. 165
into all that follow separately, then the second into all that
follow, the third into all that follow, &c,
Ex. 1. (1-+ 2aj+ 3iB«)« = 1 + 4aj» + 9aJ*
= 1 + 4iB+ lOar" + 12a;»+ 9a^.
Ex. 2. (a»+3a'6 + 3a6« + 6y
= «• + 9 a*6=» +9 a V
6a'6+ 6a*6«+ 2a«6« +6«
18a»6»+ 6a'6^+6a6'^
= a«+6a»6 + 15a*6^ + 20a»6»+15a*6* + 6a6» + 6«.
Ex. VIT.
1. Find by inspection the squares of a; - y, 3 a - 5 ft, 4 c'
+ cP, 3aj» - 23^'.
2. !Flnd the continued product of a + b, a^ + 6', a* + b*,
and a - 6.
3. Multiply mx + wy by nix - ny, and 5 a' - 3 6' by 10 a'
+ 6 6».
4. find the value of {a + b + c + d) (a - b + c - d), and
of (a -k- b - c " d) (a + b + c + d).
5. Show that (aj'+2scy+3^) (a:»-.2icy + y=) = i«*-
2a?y» + 3^.
6. Multiply 05 + 5 separately by aj + 1, a; + 2, aj - 3, a; - 5.
7. What is the continued product of aj - 2, a; + 3, aj - 6,
a; + 5?
8. K2« = a + 6+c, find the value of s (« - a) (s - 6)
9. If2« = a + ft + c + <:?, what is the value of —
(2« - 2a)» + (2« - 25)^ + (28 - 2c)« + (2« - 2c0'1
Prove that —
10. (ax + 6y)' + (ca; + c^y)'^ + (ay - 6a;)' +. (cy - c^aj)' =
(a* + 6" + c» + cf ) (a? + f).
11. I (ac - 5cf) a? + (a<£ + ftc) y P + \ (ac - bd) y - (ad
+ 6c)a:j« p= {a" + b') (c' + cf^ (ar» + y->
12. (oaj + fty + czy = (a + 5 + c) (aar* + fti/* -V ca?^ -
«6 (as - y)' - oc (a? - ;^^* - &? (y - «)'.
166 Aloe^bA.
13. (a; + a) (aj + h) (a; + c) = (a; - a) (x - 6) (a; - c) +
2 {{a + h + c)oi? + (ibcj.
14. (m + 71 + ^ + qf = (w + n)^ + (wi + pf + (w + qf
+ {n+ pY + {n + q)]+ {p + qf - 2 {in? + w^ + jp^ + ^).
15. (a^ + 6^ + (T^) (m^ + 71^ ^ ;?* + 5^) = (own- 6» + cpf
+ (aTi - 6wi + c^')^ +{ap - hq - cmf ■\- {aq + hp -^ cnf,
16. {a - 6) (6 - c) + (6 - c) (c - a) + (c - a) (a - 6) =
3 (a6 + ac + 6c) - (a + 6 + c)^
17. (a5 + y + « + a)^- (a;-y-« +a)^ = 4 (a: + a) (2/ + s).
18. {(a - 6)2 + 4a5} {(a + 6)^- 4 a6} {a* - 6* +
2a6(a2-62)}= (a + 6)' (a - 6)1
19. {a? + 6^) (c^ + (f) = (oc ± M)^ + (oc^ + 6c)l
20. 4 (a^ + 6^ (c^ + c^2) = (a + 6)^ (c + cQ" + (« - 6)»
(c - c;)^ + (a + 6)2 (c - rf)2 + (a - 6)» (c + cQ»*
2L 8 (a* + 6*) = (a + 6)* + 6 (p? - 6')^ + (a - 6)^
22. (a; - yf + (2/ - zf -{■ {z - xf ^ 1 {x - y) (y r- a) +
2 (a; - y) (a - a) + 2 (y -«)(« — a?) = 0*
23. (a + 6 + c) (i + c - a) (a + c - 6) (^ + 6 — c) =
4 aWy when a^ + 6^ = c"*
24. {2 (oaj** + 6^**)^ + (ay» - &»**)*} = (a^ + h^ {(a^ + 2^")'
bivision.
27. As in multipiicatiori^ we have fespecially three things
to attend to, viz. : —
L The signs^
We have learnt (Art. 13) that like signs give + ; and unlike
signs give - .
2. The coefficients.
Understanding here the numericcd coefficients,, it is plain
that they may be divided as ordinary arithmetical quantitiei^
3. The Utters.
As the product of the quantities a and 6 is expressed by
a3, it follows that the quotient of ab by either of the factors
a or 6 will give the remavniag iaoVw b at ob ^^^^wj^vdy.
DIVISION. 167
Thus, ah -r a or — = h, and a6 -r i or — - = a.
a
And so, osyz -r xz = y, and pqrs i- qs = pr.
We may then conclude that when the divisor is contained
as a/actor in the dvuidend, the qiwtient is fownd hy omitting
from the dividend those of its factors which constitrUe the
divisor.
If the divisor be not contained as an exact factor in the
dividend, we may then express the quotient symbolicallj.
Thus, ow V- a6 = -^.
ao
When, however, the dividend and divisor have a common
factor, it is plain that we may, as in arithmetic^ strike out of
the numerator and denominator of the symbolical quotient
this common factor.
Thus, 5 «6c * M = ^ = ^.
od d
And 16 «^'^ 10 a<«« = JJ^ = 1^.
10 aooz 5 a
28* A pOVer of a quantity is divided by any other Jpower
of the same quantity by subtracting the index of the (Uvisor
from that of the dividend, the quotient being that power of
the quantity whose index is the remainder so obtained.
1. Let the power of the quantity in the dividend be the
higher.
We have a' = aaao/a^ and c? = (wa.
aaa :._._^
Or, generally, m being greater than n, since
a** = aaa., .to m factors, and a** = aaa...to n factors,
we have —
— . - aa£(... torn factors ", ' \c i.«*« ^m— «;
a* -I- a* = rr— - — =^aaa»,Am - w) factors = a"* ***
aaa...U} n factoTa ,. _^
2. Let the power of the quantity in the dividends \^ V!&&
lower.
168 ALGEBRA*
Suppose we tave to divide a* by d.
'Weliavea*-a'= J5??l_ = J-"= L
We may, however, so express the result that it shall agree
exactly wilJi the proposition at the head of this article.
For we may conceive of a' as representing the product of
mwty and the quantity c^. We shall therefore be perfectly
consistent if we allow c? to represent the qaotimt of unity
by the quantity a*.
We shaU then have a "* = -^, and hence'we get from the
above result —
a* -5- a^ = a-' = a^-^
Or, generally, m being less than n, we have —
a** ao/a to m factors 1
a*^ ^ aaa to n factors aaa (ti - m) factors
= _ ; or, using the notation just explained,
3. Let the powers of the quantities in the dividend and
divisor be equal.
It is evident that their quotient is unity.
And SO, — = 1.
a**
If, however, we assume the principle proved in the two
cases above to hold here, we have —
^ = a«-^ = a\
a**
It follows therefore that a° = 1.
Cor. — From the above interpretation of negative indices
it follows that the same rules for multiplication and division
of quantities involving them may be applied as in the case of
positive indices.
Thus,a» X a-5 = a' x 1 = -' = a»-».
a^ a^
'Andao,a' t a~^ = a*-^-^» - a'.
sinsiojr. 169
Ex. 1. ?^ = Bxy.
Ex. 2. ^4^ = - 3aV.
Ex. 3. z2fLtii|!i::iHf*' =«._ 3a6 + 26'.
4rf 2 14 ^
+ — , or
3 y's? j/is; aj«
= |«'y^'«-' - 2y-^«-i + Ux-h-K
Ex. 5. Divide 3a« + 13a6 - 10 6' by a + 5 6.
When the divisor, as in this example, contains more than
one term, it is generally convenient to follow the method of
arithmeticallong division. Thus—
a + Bb)3a^ + Uab - 10 b\3a - 26
3 a' + 15 a6
- 2a6 - 106'
- 2a6 - 106'
Ex. 6. Divide a* + a'6' + 6* by «» + a6 + h\
a' + a6 + 6^) a* + a'6' + 6* (a' - a6 + 6'
a^ + a»6 + a'6'
- a»6 + 6*
- a'6 - a?6' - a6»
a'6' + 06^* + 6^
a'6' + a6» + 6^
It will be seen that in the last two examples care has been
taken to keep the terms of the divisor, dividend, and successive
remainders arranged according to the ascending or descending
powers of some letter. In these cases we have arranged the
terms according to the descending powers of a, and, as there-
fore follows, according to the ascending powers of 6. Want
of care in this respect will often render the operation of find-
ing the true quotient tiresome, if not impossible. The next
two examples will illustrate this point. TVie^ tc^^ \^
170 iMEfillA.
attempted first by the student^ keeping the terms in the order
as given.
Ex. 7. Divide 2 - 7 a; - 15 ar' by 5aj - 1.
- 1 + 5aj)2 - 7a; - 16a?{- 2 - 3a;
2 - 10 a;
3 a; - 15ai»
3 a; - 15 a^
Or thus, 5aj- 1)- 15 aj*- 7a; + 2(-.3a;- 2
- 15ar^ + 3a;
- 10a; + 2
- 10a; + 2
Ex. 8. Divide a;? + ^bya;*"^ + yK
Here we have the powers of a; in the dividend descending,
while in the divisor iiey are ascending. Arranging them in
the divisoi* as in the dividend,.the operation is easy. Thus —
y ^ + a^"^)af* + ^ .(a;'y - a?^ + x^
a? + aPy
- ar*y + y*
- aPy — ajy"
a^ + S/*
We shall work the next example in two ways to illustrate,
firstly, the above point again ; and, secondly, to show how the
operation may be sometimes abbreviated by the use of brackets;
Ex. 9. Divide a;* + y* + js* - 3a^« by a; + y + «.
x + y + z)9iP''3xyz + f^ + s?{a?"Xy-acz + ^''yz + z^
a? + a?y + aPz
- aPy - on/^ - xyz
- 3C^z + xy^ - 2 ocyz
-aPz —xyz-xsP
on^- ocyz + aa? + j^*
- xyz + aa? - ^«
- xyz - ] fz-y^
xs^-vy^-^^
Divisioir* 171
Or thus, InclosiDg the last two terms ot the divisor in
brackets —
- (y + «)£?- 3 ajjyis
(y'-yg-f a^g + y' + g'
It will be seen in both the above operations that we have
brought down the terms of the dividend only when the sub-
trahends indicated they were required This often prevents
much useless repetition*
fix. Vlil*
Find the quotient oi^^^
1. 28 a«6 - 7 aft» + i4 ftHy 7 6; 3 a?y^ - 12 a:y^ by
3 X7J.
2.-6 a% + 15 aV- 20 aW by - 3 ah; 4 xY + 6ar^y»
+ 4 a?y* by 2 jcy*.
3. aflc'**+ ba^y^+ cy** by »"•+*; 0^7'* + haf^'"'t/' +
cy** by of*"*;
4. 30 ar* +2 ay - 12 y^ by 5 a; - 3 y; hndby6a: + 4y.
5. l + 2a;+3aj' + 2a?+a5*byl+a;+a^.
6. 12 - 19 a; - 21 ai" by 7 a? - S ; and by 3 a> + 4.
7. aJ* - 4 a?^ + 12a^- 9y*bya;-3y; and })y a: + y.
8i a:^ - y* by aj - y; iand a:^ + y^ by a; + y;
9; cu^'*''* + adaf^i/^ + hcafy^ + Wy?J.7 W ca3^ + dy^,
10. a« + a^6» + a«6^ + ^ by a^ - a'6 -"^a6' + h\
11. 6^ + a5 + 6c + ac by a + 6 ; and a' + o6 + 6c +
ac by a + c.
12. a + (a + 6) a; + (a + 6 + c) a;^ + (a + 6 + c) ar* +
(6 + c) a^ + c«* by 1 + a; + ar* + a?.
13. o^ - pa^ + ga^ -- ^a^ + pa -^ Ihj a - \ .
172 ALGEBRA.
U. a* + 6* + cj + c«* - 2 (aV + aV + a^d^ + 6V +
fc'rf" + c'cP) - 8 oSjti by a - 6 + c + cf ; and that of this
quotient by a - 6 - c - rf .
15. Show that the remainder, after the division of a^ - pa^
+ 5x'-nc+«bya; — o, isa*- pc^ + ^ra" - ra + «.
16. Divide i«* - y* by a"' — y"*, and ai* + a?*"^ + 2 by
17. Show that the quotient of 1 by 1 + a, is 1 - a; + x*
- fl^ + <fec. cu;? infinitum.
18. Show that the quotient ofl byl — 3aj+3ai'--a:*
isl + 3a; + 6a?+10a? + 15ai* + &c.a<i infinitum.
19. Divide (a; + y)* — 2 (a? + y)* »* + 2^ by aj + y - a.
20. Divide(a-.c)»-3(a-c)*(5-cf)+3(a-c) (6 - <f)»
- (6 - e^' by a - 6 - c + rf.
Factors.
29. The ordinary method of finding the quotient of two
algebraical quantities having been explained in the last
article, we shall now proceed to show how, in certain cases,
this method may be avoided, and the quotient written down
at sight. It may be remarked at the outset, that the resolu-
tion of algebraical expressions into their elementary factors
is a subject of very great importance, and one which the
student will do well to thoroughly master.
(I.) The form a* + 2 aa; + a'.
We have seen (Art. 26) that a?+2aa; + a' = (a; + a)\
Hence the sum of tlie squares of two quantities, together
with twice the product of the quantities, is equal to the
squa/re of the sum of the quantities.
And hence, any algebraical expression, which can be thrown
into the form (a:? + 2 oa; + a^), is of necessity a perfect square.
Thus—
a? + 6 a; + 9 = ar^ + 2 (3) a; + 3^ = (a: + 3)2.
a» - 10 ofi + 25 y = a' + 2 a ( - 5 J) + ( - 5 J)* =
(a - 5 h)\
16a*aj»- 56 a% + 49 6y= i^axf -^^ 2(4a«)(-7iy)
^ (^ - 7 «y)« = (4 oa: ^ 7 hy)\
FACTORS. 173
3 o^V - 6 ahcxy + 3 a(?f = 3 a (6 V - 2 haty + cV) =
3 a{(&c)» - 2 (6a?) (cy) + {cyf] z: 3 a {bx - cy)\
(II.) The form a* - 61
We have seen (Art. 26, 11.) that a'- 6' = (a + h) (a- 6).
Hence the difference of the squares of two quantities is
equal to the product of the sum and difference of the
quantities. Thus —
aV - 6y = {axy - (6y)' = {ax + hy) {ax — by),
**-y*=(aO"- (2/^)' = (ic» + 3^(0? - 2^
= (cB* 4- 3?") (a; + y) (aj - y).
a«+6«-c»-cP+2a5-2cc^
= (a* + 2 a6 + 6') - (c« + 2 cd: + <£2)
= (a + hy - (c + c^)»
= {(a + 6) + (c + i)} {(a + 6) - (c + rf)}
= (a + 6 + c + c?) (a + 6 — c - cf).
aj* + a*y* + y* = (aJ* + 2 «y + ^) - ^1^
= (a* + ^)' - a:?y2
= {(aJ* + ^) + ay} {(aj" + y*) - »y}
= (a? + xy + y*) (as* - a?y + ^).
(in.) The form a;* + j»a; + g'.
This form evidently includes both the preceding, for the
first form — viz., a? -{• 2ax + a' is included, since q may be
the square of half p ; and the second is included — ^viz.,
a? - a*, since we may have p = o, and q a negative square
quantity.
Now, the resolution into elementary rational factors of the
quantity ai* + ^aj + 5^ is not always possible ; but, since
(Art 26, III),
a? + (a + 6) a: + a6 = (a; + a) (a; + b),
we have the following rule, when the quantity admits of
resolution.
BuLE. — ^If the third term q of the quantity a? + px + q
can be broken up into two factors, a and b, such that the
sum of these will give the coefficient of x, then the
elementary factors of :? + ^a; + 5^ are x + a and x + h.
Thus, a? + 7aj + 12 = (a; + 3) (aj + 4); for the 'product
of 3 a©^ i ^ 12j and their fium is 7, the coeffideii\» ^i v.
174 ALGEBRA.
irf - oj - 30 = (aj - 6) (aj + 5); for the product of - 6
and 5 is ~ 30, and their sn/m is - 1.
And so, a^ - 18a; + 32 = (a? - 2) (a - 16),
And a* + 3 oft - 108 6« = a* + a2 6 - 9 6) a +
(12 b) (- 9 6) = (a + 126) {a - 95).
(IV.) The form aa^ + hx + c.
This is the general form of a trinomiaL The following
remarks, though equally applying to each of the three pre-
ceding forms, are especially intended to be practicaUy
applied to trinomials not included by them.
The above form will include such expressions as the fol-
lowing:— 20 a:» + 11 a? - 42, 6 aj* - 37 a; + 55.
It is evident that the product of the first terms of the
factors will be the first term of the given trinomial, and that
the product of the last terms of the factors will be the third
tenn of the given trinomial.
And, further, when the third term is negative^ the last
term of one factor must have the sign + , and the last term
of the other the sign -^ ; but, when the third term is
positive, the last terms of the factors must have the same
sign as the middle term.
Thus, 12 ar» - 31a; - 30 - (4a; + 3) (3a; - 10).
Here the factors of 12 a? are either 3 x and 4 a;, 6 a; and 2 x,
12 a!; and a;, and the factors of 30 either 5 and 6, 3 and 10,
2 and 15, 1 and 30 ; and we must give a + sign to one of each
of these latter pairs, and a - sign to the other. It is easily
found on trial that, in order to obtain - 31 a; as the middle term,
the factors of the trinomial must be 4 a; + 3 and 3 a; — 10.
SowehavelOa* - 41 a5 + 21 6^ = (5a - 36) (2a - 76),
and aca? + {ad + he) xy + hdy^ = {ax + In/) (ex + dy),
(V.) The forms of* + ^ and as" - ^.
We shall show in the next article that a rational in-
tegral algebraical expression, involving x, contains as - a as
a factor when it Vanishes on substituting a for as.
Hence, of* + y^ and af* - ^ must each vanish on putting
y for X, if they contain a; - ^ as a factor, n being an integer.
The former becomes ^ + ^ or 2 ^, and the latter ^ -
y* or 0. We therefore conclude that —
pf* + ^ does not contaiax - ^ «&^i«ji:^, and that —
FACTORS. 175
rt" - y does contain x - y as & factor, whether n be
even or odd.
Again, on the same principle, they must each vanish if
they contain a; + ^ as a factor, on putting - y for x.
The former becomes ( - yY + y", which vanishes when w isodd,
andthelatter becomes ( - y)" - y^jwhich vanishes when wis even.
Hence we conclude that —
af* + y" contains aj + y as a factor when n is odd, and
a^ - i/^ contains x + y aa a, factor when n is even.
Now, iie quotient of either of these quantities by a; + y or
a, - y can in any particular case be found by long division.
We thus find that —
ix^ + j/^ = (x + y) (x^ - a^y + otPy^ - xi^ + y*),
a^ - j/* == {x - ySla^ + a^y + xy^ -¥ f),
a? -y^ = (x - y) (oc^ + xy + 2^).
The law of formation of the co-factor in each case is easy
to see; and if we may assume this apparent law as generally
true, we may conclude that, when an algebraical quantity is of
the form af + y" or 05" - y", and it contains x + y or x - y
as a factor, the law of formation of the co-factor is as follows: —
Law of Formation of Go-Factor.
1. The terms are homogeneous, and of dimensions one
degree lower than the given expression, the power of a; in the
first term being n - 1, and diminishing each successive term
by tmUy; and the power of y increasing each successive term
by tmity, and first appearing in the second term.
2. The coefficient of every term is unity.
3. The signs are alternately + and - , when a: + y is the
corresponding elementary factor ; and are all + , when x - y
is the corresponding elementary factor.
Ex. 1. a*^ -I- 32 = a» + 2^^ = (a + 2) (a* + a» • 2 + a*- 2^
+ a •2» + 2*) = (a + 2) (a* + 2 a' + 4 a« + 8 a + 16).
Ex. 2. »• - 6« = {ay - (by = (a« + h^ (a» - h') =
a ^h)(o?-ah + V^y^a - h) (a^ + a6 + 6^ ^ {a + b){a- h)
a^ ^ ab + }^) {a^ + ah ^ h%
80. The remainder of the division of a rational integral
function of is. by yi - a may be found by putting a for x in
the given fimction^
t
176 ALGEBRA.
Def. — ^A fimetion of a; is an algebraical expression in-
volving X ; and a rational intend fanction of a; is an expres-
sion of the fonn aaf^ + to""^ + Ac + sx + t, where all
the powers of x are integral and positive.
Lety (x)* be a rational intend fanction of x, and suppose
Q to be the quotient, and B the remainder on dividing the
function by a; - a.
Then, evidently —
Q {x - a) + B = /{x) identically.
And this identity must hold for all values of x, and
therefore holds when x = a.
In this case we have Q{a - a) -i- E = /(a)
+ E =/(a)
orE =/(»).
Now/ (a) is the result of putting a for £ in the given fono-
tion, and is, as we have just diown,^ the remainder on
dividing the given function by oi; - a.
Cob. 1. When there is no remainder, we must, of course,
have/ (a) = 0. Hence, a given rational integrsd function
of X vanishes when a is put for oi;, if it be divisible by a; - a.
Ex. 1. The remainder, after the division of 2 a^ — 5 a^ +
6 a; + 7 by aj - 2 is 15.
For, putting as = 2, we have —
2a:»-5a* + 6a; + 7 = 2- 2» - 6 •2« + 6 • 2 + 7 = 15.
Ex. 2. The function—
a?-2a;? + 5a; — 62is divisible by a; — 4,
For, putting a; = 4, we have —
aj»-2a? + 5a;-.52 = 4»-2-4* + 5-4-62 = 0.
OoB. 2. Any rational integral function of a; is divisible by
a; — 1, when the sum of the coefficients of the terms is zero.
For, putting a: = 1 in the given function, it is evident
that it is reduced to the sum of its coefficients, which sum
must be zero if the function be divisible by a; — 1.
Ex. Each of the following functions is divisible by
aj - 1, viz.: —
3aJ* + 7aj»-ar' + 12aj-21, 5aj'-2aj-3,
(a-6)a^ + (b — c) X + {c — a),{a-{-hya?—iabx^(a-hy.
* The expression f {x) must not be considered to mean the product
of/ and Xf but as a «ynilK)l uaed lox <joii^ w«a<»^
FACTORS. 177
Cor. 3. Any rational integral function of a; is divisible by
a? + 1, when tlie sum of the coefficients of the even powers of
X is equal to the sum of the coefficients of the odd powers.
(The tenn independent of x is always to be considered as the co-
efficient of an even power).
Let oof* + oof*" ^ + <kc. + ra::^ + «B+<bea rational in-
tegral function of x.
Put a; =x - 1, then we have, if the function be divisible
by a; + 1 —
a(-l)«+6(-l)»»-i + &c. + r(-l)« + «(-l) +< = 0.
Suppose n to be even, then evidently (—1)"= (-1)
( — 1) ( — l)...to an even number of factors = + 1.
And so ( - l)«-i = ( - 1) ( - 1) ( _ l)...to an odd
number of factors = — 1 ; and so on.
Hence we get a - h +&c. +r — «4-^=0, and this
must evidently require the condition that the sum of the
positive quantities is equal to the sum of the negative, and,
therefore, that the sum of the coefficients of the even powers
of a; is equal to the sum of the coefficients of the odd powers.
And a similar result will follow if we suppose w to be odd,
Ex. Each of the following functions is divisible by a; + 1,
viz.: —
a'+5a» + 7aj + 3, 5a:5_4^^3^_2a;- 1,
aV - (a + 1) (a + 2) ar* + 2 a: + 3 a + 4, ^ + ((? + r) a.^
+ (^ + r) a; + ^.
Ex. IX.
Ilesolve into elementary factors —
1. a» - 9 a', 16 2^ - 25 «*, 24 a« - 54 ft^ 8 a» - 27 if,
2. aJ* - aJ2/*, a* - h\ a;?/* + x% 2 x^z - 8 xfz^.
3. a* - 4 6*, a^ + a?f-¥i/,a^^2 aW + h\ a» + t^ _ c* +
2 ah.
4. a* + h^-i?^d:^ + 2a6-2c(Z, a^ - 52 _ ^2 ^ cP + 22>c
+ 2 flkf, a» - (6 - c)\
6. {x + 7)« - (a; + 2)\ (x + 5)^ - (a; + 2)«, (2a + 6)«
- (a - c)l
6. {a? ^ y^y + 4:{a^ + x'f + y*) r,^f, {a? + ff -
5(0* + ffni?f + 4a?y.
178 ALOEBBA.
7. a? - 3aj - 70, ar' + 11 aj + 10, a= - 15a6 + 66 h\ a?
- 4aj- 192.
8. araP + abxij - 42 by, 3 aa?- 24 aaj - 60 a, 24 oc -
o cUj A" ox?
9. 6aj= - 11a; -35, 8ar* + 6a; - 135, 18ar» - 21a; - 72,
20ar» - 11a; - 42.
10. Zo^y + l^Q^y" + 3ay», 20a:' + 12aai* + 25 6ar +
1 5 aibx, 7)fi?3t? + {piq + mp) x + pq.
Write down the quotient of —
11. aJ* - 16 by a; - 2, 33;* + 96 by a; + 2, a:" - 27 by
ai«-3.
12. (a + ft)» - (c + cf)' by a + ft + c + J, a' - 6^ + c*
Find the remainder after the division of—
13. a^ + po? + qaP + rx + « by a; - a, x* + a* by
a? - a\
14. ar» - 5a;^ + 7a; - 9 by a; + 3, a;* - 3a; + 7 by a; - 2.
Show that —
15. 5 a;* -3a;'+ 7a;'-8a;-l is divisible by a; — 1.
16. 2a;* - 3ar» + a;» - 7a; - 13 is divisible by a; + 1.
CHAPTER III.
Involution and Evolution.
Involution.
31. Involution is the operation by which we obtain the
powers of quanl^^ties. This can of coiu*se be done by multi-
plication, but the results obtained by the actual multiplication
of simple forms enable us to develop without multiplica-
tion more complex forms. As the subject requires the aid of
the Binomial Theorem, we shall here show how to develop a
few only of the more simple expressions.
- 32. The power of a Biivglo tom^a OtAaMi^ \i»Y taiaing the
J>'VOl.UTION, 179
coefficient of the term to the power in question, and muUiplt/-
ing the exponents of the letters of the term by the exponent
of the power in question.
Thus, {a?f = a' X a^ = ^3 + 8 ^ ^3 X 3^
{cC^Y = a** X a** X a*^...,.. tow factors = (i»» + »» + »+... to nteruw
= a"^.
And so, (4a«6V)' = 43a2x36'^V^3 ^ 64a«&Vl
33. Developmentof the^/wr<i,ybwr<A,andj^i5/* powers of a + h.
We know (Art. 26, 1.) that {a + hf =^ a" -¥ ^ah •¥ b\
.-. (a + by = (a' + 3 a6 + 6^ (a + 6) = a' + 3 a«6 + 3a6» + 6J;
also, (a ^ by = {a^^ Sah^ + ab^ + b^) la + b) = a^ +
4a'6+ QaW + iaW ^ b';
and {a + 6)' = (a* + 4 a^6 + 6 a^ft^ ^ ^ ab^ + j*) (a + j)
= a» + 5 a*6 + 10^862 + lOa^fi^ + 5 a6* + 6^
The following law of the formation of the terms is evident :—
Zaw of Forrtvation of Terms,
1. The first term contains a raised to the given power, and
the power of a decreases by unity in each successive term,
while the power of 6 (which first appears in the second term)
increases by unity in each successive term, till it reaches the
power of the given quantity,
2. The first coefficient is unity, and the coefficient of any
term is found by multiplying the previous coefficient by the
exponent of a in the previous term, and dividing the product
by the niunber of terms hitherto developed,
Ex. 1. {2x + 32/)' = (2a;)8 + Z{2xy{^y) + 3(2 a) (3^)2
+ {Zyy = Sic' + ZQx'y + 54a^ + 272^.
Ex. 2. (a + 6 + c)« == (a + 6 + c)' = (a + hy +
3(a + bye + 3 (a + ^>) c* + c» = (a« + 3 a^fi + ^ab^ + ¥)
+ 3 (a« + 2 a6 + 6*) + 3(a + 6) (r» + c^ = a» + «»» + c^
+ 3a^6 + 3a'c + 3a6' + Zbh + Zax? + ^b(? + ^dbc.
(In the following examples the above }aw may be damned as
generdUy ttxie.)
Ex. X.
1, Vind m valtm of (»**»)» ( - 3 ffl6^», \^-<i%Y, V- «?y^
180 ALGEBRA.
'Expand —
2. {a + 3h)\ (2a + h)\ (a - by, (3a - 46)»
3. (2m + l)^ (5 a; + 2)', (3a - 4c)^ (-a - 6)».
4. («» + re + 1)*, (3 a - 6 + 4 c -</)'>(« + 2 6 - c)»
(3a+ 36 + 3c)».
5. (1 + Jc - a*)', (oa; + 6y + «)', | (a + 6)x-(c+ (f)y I .
6. (1 + xf, (1 + a; + a»)*, (a + &c + car)*.
7. (oaj - Jy)», (3 a; + y)»(3a; - y)', («* + icy + jr)' (^ - y)'-
8. (a^ + ar^y» + 2^)M^+ y)M« - y)',
j(a + fc)«-4a6J'-j(a - 6)« + 4 a6 | .
Simplify —
9. (a + 6 + e)' - 3 (a + 6 + c)»c + 3 (a + 6 + c)c' - c*.
10. (a - 6)» + 3 (a • 6)2 (6 • c) + 3 (a - 6) (6 - c)H
(6 - «)••
11. (1 + a: + 3a:» + Saf)^ + (1 - aj+3ar^ - 3a:^l
12. j (aJ + y)' - (a^ + y») [ ' - 27 a:'^' (a? + y»).
Evolution.
34. Evolution is the operation by which we obtain the
roots of quantities.
Since the squa/re or second power of a' is (a*)* or a\ we call
(Art. 17) a' the sqwi/re root or the second root of a*.
And so, since the cfvihe or <Air<i power of a' is (a^)' or a*, we
call c? the cw5e root or the <Air<i roo< of a'.
So, generally, since the nth power of a^ = (a"*)" = a"*", we
call a"* the wth root of a*^. *"
Thus, we have n/o* = a', 4^' = a^ ->/a*^ = a"*.
Hence, in the case of quantities consisting of a single letter
with a given exponent, when the given exponent contains as
a factor the wwmheft indicating the root, we must divide the
given exponent by this number, the quotient being the
exponent of the root,
(We sball see farther on that this rule holds when the given exponent
U not 00 dimibkf the root in thia c«AQ\>^m^^^ « ffnrd).
SQUARE ROOT. 181
35. Since the product of an even, number of negative
JaxtoTS must give a positive result, and the product of an odd
number of negative factors a negative result, it follows that —
I. When the root is indicated by an even number —
1. The root of a positive quantity may be written either
with a + or - sign.
Thus, >^d^ = ±3 a, 4^166* = ±2b.
2. The root of a negative quantity is impossible.
Thus, V - a^ V - a^b^\ &c., are impossible quantities.
II. When the root is indicated by an odd number, the root
has always the sign of the given quantity.
Thus, 4^ - 27 6« = - 3 ftS ^32 a^y = 2 ar'y.
(It may be remarked that the theory of impossible quuintities forms an
important branch of Algebra, which the student cannot yet enter
upon. According to that theory, all quantities have as many roots as
the number indicating the root.)
Square Root.
36. We shall now develop the method of finding the
square root of a given quantity.
Ex. 1. Find the square root of a^ + 2 a6 + b\
We know that a^ + 2ah -{■ h^ = (a -{- by.
Hence, ct + 6 is the square root of a' + 2 a6 + 6' or a*
+ (2 a + b)b.
Now, it is evident that the first term a of the root is the
square root of the first term a' of the given quantity;
and if this term be subtracted, there remains 2 ah •¥ b\
from which to determine b the second term of the root.
Now, b is contained in 2 a6 + 6^ or (2 a + 6) 6 exactly
(2 a + 6) times. Hence it follows that the second term of
the root is foimd by dividing the remainder by twice the
first term of the root, and, if we wish to arrange our work
in a way similar to long division, it is evident that we
first take for our divisor 2 a + 5, that is, twice the first
term of the root added to the second term, which mul-
tiplied by b the second term, and subtracted, leaves no
remainder.
182 ALOEBftA.
Thus the whole process may be arranged as follows : —
a?
2,ah -¥ y
Ex. 2. Find the square root of a^ + 2 a6 + 6^ + 2 ac
+ 2 6c + <r*.
Now we know (Art. 26) that —
a^ + 2 a5 + 6» + 2 ac + 26c + c= or (a + 5)2 +
2 (a + 6) c + c^ = (a + 6 + c)\
And if we compare the form {a -k- Vf •¥ 2 (a + 6) c + c^
with the form a^ + 2 a5 + 6*, it is evident that, having ob-
tained as in the last example the first two terms a + 5, we shall
by continuing the process obtain the third term. Thus —
a^ + 2 a6 + 6^ + 2 ac + 2 5c + c'(a + 6 + c
2a + 5 2a5 + 62
2a5 + y
2a + 25+c 2ac -{■ 2bc -h c^
2 oc + 2 6c + c^
We may deduce from the above examples the following
general rule : —
Rule.
1 . Arrange the terms of the ^ven quantity according to the
ascending or descending powers of some letter, and take the
square root of the first term for the first term of the quotient.
2. Subtract the square of the quotient, and bring down
the next two terms of the given quantity.
3. Double the quotient, and place the result as a trial
divisor; then, (lividing the first of the terms brought down
by this trial divisor to obtain the second term of the root,
add the quotient so obtained to the first term of the root,
and also to the trial divisor, to obtain a complete divisor.
4. Multiply the complete divisor by the second term of
the root, and subtract the product, as in long division, from
the terms brought down.
5. If there be any remainder or more terms to bring down,
double tb^ whole quotient fox a tnai di\iaor, and divide the
SQUARE ROOT OF NUMERICAL QUANTITIES. 183
remamder by the new trial diyisor, to obtain the third term
of the root; and so on.
Ex. 6. Find the square root of 1 - 4a; + lOar^ - 12a? + 9a^.
Thq terms are here arranged according to the ascending
powers of X. Then proceeding according to rule, we have —
I - ix + lOa?- 12cc» +9aJ*(l-2a; + 3a:»
1
2-2a; - 4c x -h 10 a?
- 4 a; + 4 a;*
2-.4a; + 3a? 6ar» - 12a;» + 9a^
6a^ •- I2a? + 9a; *
The student will observe that twice the quotient is most
easily obtained by bringing down the previous complete
divisor with its last term doubled.
Square Root of Numerical Quantities.
37. It is easy to apply the above method to numerical
quantities.
Since V = 1, 10« = 100, 100«= 10,000, 1,000«= 1,000,000,
<&c., it is evident that the square roots of numbers having less
than three figures must contain one figure only ;
That those having not less than three and less than five
must contain two figures and two only ;
Those having not less than five and less than seven must
contain three figures and three only ; and so on.
Hence it follows that, if a dot be placed over the units'
figure, and over every alternate figure to the left, the num-
ber of dots will give the number of figures in the square root.
Thus, the square roots of the numbers 141376 and
• • • •
1522756 have three and four figures respectively.
In the number 141376 we call 14, 13, 76 respectively the
• • • •
first, second, and third periods. So in the number 1522756,
the first, second, third, and fourth periods are respectively
1, 52, 27, 56.
It is evident that the number of periods correspond to the
number of figures in the squa^ root, and it will be seen that
the figures of each period are used in the operation for tk^i
oorrespondu2^ jBgure of the root.
184 ALG£BItA.
Ex. Find the square root of 565504.
Pointing off the number, we find the first period to be 56.
Now, the greatest square in 56 is 49, and the square next
greater is 64. Hence the number lies between ihe square
of 700 and 800 ; and, following the algebraical method, 700
will be the first term of the root
The operation will stand thus — a b e
565504 (700 + 50 + 2 = 753
49000 = a«
2a + 5 = 1400 + 50 =1450 75504
' 72500 =^ 2ah + P
2 a+25 + c = 1400+100+2 = 1502 3004
3004 = 2ac + 2hc + (^
Or, omitting the useless ciphers, and bringing down odo
period of figures at a time, the operation will stand thus —
565504(752
49
145
756
726
1502
3004
3004
Ex. 1. Find the
square root of 6091024.
6091024(2468
4-
44
209
176
486
3310
2916
4928
39424
39421
Ex. 2. Find the
square root of 83521.
83521(289
4
48
435
384
569
5121
5121
SQUARE ROOT 01* A DECIMAL. 185
It will be observed that the second remainder, 51, is
greater than the previous complete divisor, 48, and it might
be supposed, therefore, that the second figure in the root
should be 9 instead of 8.
Now, the square of (a + 1) exceeds tho square of a by
2a + 1.
Thus, (a + 1)2 - a= = (a^ + 2 a + 1) - a' = 2 a + 1.
Hence it follows that, so long as any remainder is less than
twice the corresponding number in the root + 1, we may be
certain that we have taken the figure of the root sufliciently
large.
Thus, since the remainder is less than 28 x 2 + 1
or 57, we may be certain that 8 is the correct figure and
not 9.
Square Root of a Decimal.
88. It is evident that the aqua/re of any number containing
one, two, three, &c., decimal figures, will contain two, four,
six, &c., decimal figures respectively ; and, hence, conversely,
every decimal considered as a square must contain an even
number of decimal figures, and its square root must contain
half this even number of figures. It will then be necessary
to add a cipher when the given number of decimal figures
is odd.
Further, since decimals and integers follow the same system
of notation, it is evident that if a dot be placed over the
uvM figure of the given number, the pointing off may be
performed with regard to the integral part exactly as in
integers, there being no necessity to point off the decimals,
only taking care to bring them down in pairs, and putting a
decimal point in the quotient when the first pair is brought
down.
And again, if an integer be given which is not a perfect
square, we may, by affixing to the right of it a decimal point
and an even number of ciphers, gradually approximate to the
square root as nearly as we please.
186
ALGEBRA.
Ex. 3. Find the square root of l-8225r
23
265
1 •8226(1 -35
1
82
69
1325
1325
Ex. 4. Find the square root of 247 to four places of
decimals.
Instead of adding a decimal point and eight ciphers to
the right of the given number, we will proceed in the
ordinary way till we arrive at a remainder. Then putting a
decimal point in the quotient, we shall add two ciphers to
this and each successive remainder.
25
307
3141
31426
314322
247(15-7162
1
147
125
2200
2149
6100
3141
195900
188566
734400
628644
105756
89. It will be shown hereafter that when n + 1 figures of
a square root have been obtained by the ordirw/ry method, n
figures more may be obtained by dMding the remainder by
tlie numh&r forrmd by taking twice Hie quotient cUready
obtained, provided tliat the whole numJb&r of fibres in the
root w 2 n + 1.
Ex. Find the square root of 29 to six places of ded-
maJs.
Cube nooi*.
187
The square root required will evidently contain seven
figures. We shall therefore find the first ybwr figures by the
ordinazy method, and the other three by the above method.
Thus—
29(6-385164
25
103
400
309
1068
9100
8544
10765
55600
53825
10770
1 7750 then by division,
10770
69800
64620
51800
43080
Ans. 5*385164.
8720
Cube fioot.
40. We will next develop the method of finding the cube
root of a quantity.
Ex, 1. Find the cube root of a> + Sa^b + Sah^ + hK
We know (Art. 33) that {a + ft)* = a* + 3 a^h +
3 oft" + 6*. Hence a + 6 is the cube root of a'* + 3 a^h +
3aft»+ftl
We see then that, the quantity being arranged according
to the powers of a, the first term a of the cube root
is the cube root of the first term of the given quantity;
and if this term a* be subtracted, there remains 3a^6 +
3aft» + 6».
We see again that if this remainder be divided by 3 a', its
first term gives h the second term of the root, and, further, if
it be divided by ft, we get 3a^+3aft + ft'asa quotient.
If we wish therefore to arrange the whole proc^^'am^iiNi^"^
\66 Al^fififtA.
Bimilar to ordinary division, it is evident that we must
write as a divisor 3 a' + 3 a6 + 6^, in order that after
multiplication by the quotient figure b we may obtain a
quantity which when subtracted shall leave no remainder.
The operation will then stand thus —
a« + 3 a^ft + 3 ah^ + h^{a + b
a'
3a' + 3a6 + b^ ^a'b + 3ab^ + 6*
Sa'b + SaP + y
•
We call 3 a^ the trial divisor, because by means of it we
search for the second term of the cube root. Having ob-
tained this second term, we then form the complete divisor
^a" + Sab + b\
Ex. 2. Find the cube root of Sic' - 36iB'y + Hxif - 272/*.
8cc'-36ar'y + 54a^- 272/^(2a; -3y
Sa^
12ar'- 18i»y + 92/* - 36 ar»y + 54 icy* - 273/^
-36ar'y + 54a^-27y»
Explanation. — We find the cube root of 8 aj* to be 2 aj.
This is then the first term of the quotient, and corresponds to
a in the previous example. We now require 3 a' for a trial
divisor. This, of course, = 3(2 a;)' = 1203*. Subtracting the
first term of the given quantity and dividing the first term
of the remainder by this trial divisor, we obtain - 3 ^ for
the quotient. This forms the second term of the root, and
corresponds to 5 in the last example. We now easily obtain
3ab and b\
Thus, 3a& = 3(2a;)(- 3y) = - 18a;y, andft'* = (-3y)^
= 92/^.
Hence, the complete trial divisor, corresponding to 3 a' +
Sab + 6* in the last example, = 12a^ - 18a3y + 9^.
Multiplying now by - 3 y the quotient, we obtain
- 36 a^y + 54 x^ - 27 3/^, which subtracted leaves no re-
mainder.
Hence, 2x '^ Sj/^iBtliocub^ioOu*
CUBE BOOT OP NUMERICAL QUANTITIES.
189
I
CO
I
+
o
I
CO
+
o
I
O GO
I +
CO CO
+
CO
CO
CO
I
>
CO
+
00
I
<M
CO
>
CI
I
*^>
+ I
<M
eo
CO
I
>
CO
o
+
00
'^
I
<M
I
CO
I—*
+
(M
I
<M
I—*
CO
II II II
I I
+ +
CO CO
<n
I
1|
n3 •^'bp'^ ^
0)
.a
SP,S"*
W O S fli <*^
+
^ca n2^2 4.
^a5
s
(d
3
•c
O
O
•>^
a
U2
^ 9
^^ ss
190 ALGEBRA.
Ex. 1. Find the cube root of 262144.
262144(60 + 4 = 64
216000
Sa^ = 10800 46144
Zah = 720
y 16
Za^ + 3a6 + 5' = 11536 46144
Explanation. — Pointing off the given number, we find
the first period to be 262, and that tiie cube root consists of
two figures. Now, the greatest perfect cube in 262 is 216,
which is the cube of 6. Hence, the given number lies between
the cubes of 60 and 70 ; and following the algebraical method,
60 will be the first term of the cube. This, we see, corre-
sponds to a in the algebraical method.
We first then subtract the cube of a — ^viz., 216000, which
leaves as a remainder 46144.
We now write down Sa^ or 3 (60)^ = 10800, which is the
trial divisor for determining b. Dividing then by this value of
3 a^, we find 6 = 4, which is the second term of the cube root,
We next obtain 3 a5 = 3 (60) (4) = T20, and 6* = 4^ =;= 16,
and so by addition we get 3 a* + 3 ah + 6* = 11536, which
is the complete divisor. Multiplying this then by the
quotient figure, we subtract the product, and, there being no
remainder, we find the cube root to be 64.
We may omit the useless ciphers in the above operation,
if, remembering the local value of figures when nimibers are
expressed in ordinary notation, we take care to place the
right-hand figure of the value of 3 ah one place to the right
of the corresponding figure of the value of 3 a^ ; and also to
place the right hand figure of h^ one figure further still to the
right.
The operation will then stand thus —
262144(64
216
3
X
62
^
108
3
X
6x
4
^
72
4'
=
16
46144
11536 4^U4:
r
CUBE BOOT OP NUMERICAL QUANTITIES. 191
«
Ex. 2. Find the cube root of 102503232.
102503232(468
G4
3x4'
3x4x6 =
62 =
•
48
72 ^
36 f
5556 4
36 J
38503
33336
5167232
3 X 46»
3 X 46 X 8 =
8^ =
6348
1104
64
645904
6167232
Explanation. — ^The first two figures of the root are ob-
taihed as in Ex. 1 . We then treat ^e number they form, viz. ,
46y as corresponding to a in the algebraical models omitting
useless ciphers. Obtaining then 3 a^ or 3 x 46^ = 6348, we
find 8 to be the next figure of the root. Then writing under
this, 3a6or3 x 46 x 8 = 1104, and afterwards b^ or 8^ = 64,
taking- care as to the positions of the right-hand figures, and
adding, we get 645904 as the complete divisor. Then as before.
Kehark. — It is unnecessary to be at the trouble to find
the value of 3 x 46^ by ordinary multiplication. For re-
ferring to the algebraical model, and writing here the succes-
sive terms of the complete divisor, and adding, we have — -
3a«
3ab \
^ ( If we now again write down P
Sum = 3 a^ + 3a6 + b^ j under this sum, and then add up
6* ) tJie last four lines, we get —
3a2 + 6a6 + 36^ or 3 (a' + 2a6 + 5') = 3 (a + b)\
This is three times the square of the first two terms of
the root.
It therefore follows that, if , as in the above example, after
completing the operation for finding the first two figures of
the cube root, we write imder the complete divisor just ob^
tained the value of the square of the second figure, 9Jiid \>Vi%\v
add iogeiiher the last four linea thus obtained, "w^ ^^\» \jto<^
192 ALGEBRA.
times the square of the quotient for a partial divisor by which
to determine the next figure of the root.
The four lines to be added are in the above example
bracketed. This method will be found to materially shorten
the work, for it may be similarly applied to find the trial
divisor when the cube root consists of any number of figures
Cube Boot of a Decimal.
42. We know that the cuhe of any number containing one,
two, three, <kc., decimal figures will contain three, six, nine,
&c., decimal figures respectively, and hence, conversely, every
decimal considered as a cube must contain a number of
decimal figures which is a multiple of three, and the number
of decimal figures in the cube root must be one-tliird of the
number contained in the given cuhe. It will then be necessary
to add ciphers when the given number of decimal figures is
not a multiple of 3.
And by continuing the reasoning of Art. 37, if a dot be
placed over the imits' figure and over every third figure to
the left, it will be sufficient to bring down the decimal figures
three at a time, putting a decimal point in the quotient when
the first three are brought down.
And further, if an integer be given which is not a perfect
cube, we may proceed in the ordinary way till we arrive at
a remainder, and then, putting a decimal point in the quotient,
by affixing three ciphers to this and each successive re-
mainder, approximate to the cube root as nearly as we please.
Ex. 3. Find the cube root of 395446904.
X
395-446904(7-34
343
3 X 72 = 147
X 7 X 3 = 63 \
32 = 9(
52446
15339 f
46017
9)
6429904
3 X 732 = 15987
73 X 4 = 876
42 = 16
^429904
CUBE ROOT OP A DECIilAL.
193
43. We shall show farther on that when n + ^Jigurea of
a cube root have been obtained by tJis ordina/ry method, n
figures more may be obtained by dividing the remahider by
the next trial divisor, provided tliat the wlwle number of
figures in the root i9 2 n + 2.
We may apply this principle with advantage when we re-
quire the cube root of number to a given number of decimals.
Ex. Find the cube root of 87 to ^yq places of decimals.
The required cube root will evidently contain 6 figures,
and since 6 here corresponds to 2 9i + 2 above, it is evident
that n = 2, Hence, we shall find 4 (that is, n + 2) figures
by the ordinary method, and then 2 more by division.
The operation will stand thus —
87(4-43104
64
3 X 42
3x4x4
4?
=
48
48 ^
16 (
23000
5296 (
21184
3 X 442
3x44 X 3
3«
II II II
II II II
16;
5808
396 ^
584769 {
9;
1816000
1754307
61693000
3 X 443«
3 X 443 X 1
P
588747
1329 -N
58887991 C
58887991
58901283
280600900
235605132
Ans. 4-43104.
44895768
Ex. XI.
Find the square roots of —
1. 4ar'2^a», 16aV, aj* + 2aV + a\
194 ALGEBRA.
3. 25 a^ - 30a'6 + 19 a^6« - 6a6» + h\
4. 1 ^ 4a: + lOar^ - 20a» + 25aj* - 24aj» + 16rB*.
5. a* + 2a5a; + (2ew + 5'')a* + 2(at^ + hc)a? + (2 5c^
+ c2)i«* + 2cda? + <^aJ«.
6. aV« - 6a«aj^-i + 17a=aj2«-2 _ 24rtar^-» + 16 ar"-^.
7. aj^ + 2 + aj-^ drx-^ - 2 + rt-^^^.
8. 9ar^« - Sa^a;** + 25a« - SOoa;'" +^ + 5al
4
Find the square roots of —
9. 1296, 6241, 42849, 83521.
10. 10650-24, -000576, -1, ^v
11. VI 7, VI -5. -j^^^^ ;/^-ri-
Give the values correct to four places of decimals of —
12. i^^HL^, V5W2 ^ ,^2 3.j,jg „,
V93 ^5 - V2 VlO + 2
Pind the cube roots of —
13. 8a»5V2, U^n^f.a!' + 6 a»6 + 12a52 + 851
14. a:^^ + 9a;^° + 60" -- 99 a* - 42 a^ + 441a? - 343.
15. a? + 3ar*y + 3 ary* + 2/^ - 6ca^ - 3 car* - 3cy^
+ 3c^a; + 3cV - c'.
16. ar» + a:-' + 3 (aj + aj-^), a?y-^ + 3 a?2^-« + %xy'^ + 1.
Find the cube roots of —
17. 5849513501832, 1371-330631.
18. 20-346417; -037, tYtV
Give the value of the following correct to four places of
decimals : —
V5\12 + v^-03375 1
19.
4^80 - 4^-bT ' V^ + V2 + 1.
.^ IfWh + ^'0 4 . >/5 + 2
20. ►- , : ir===~- of = •
>vA05 + V-04 7
OBEATEST COMMON HEASUBE. 195
22. Vll + 6^2, If Q + 15 73.
93 J^ 6 + 275 V5"+ 1
" ' 4 16 ' 4 • _
24. a' (6 - c) - h^(a - c) + c'(<i - J), where a = Vl*2,
CHAPTER IV.
GBEATEST COMMON MEASUBE AND LEAST COMMON MULTIPLE.
Greatest Common Measure.
44. In Arithmetic (page 24) we defined the G.C.M. of two
or more numbers as their highest common factor. In Algebra
the same definition will suffice, provided we understand by
the term highest common factory the factor of highest dim&nr
sums (Art 18). This, it need hardly be remarked, does not
necessarily correspond to the factor of highest numerical value.
45. To find the G.C.M. of two quantities.
Rule. — ^Let A and B be the quantities, of which A is not
of lower dimensions than B. Divide A by B, imtil a re-
mainder is obtained of lower dimensions than B. Take this
remainder as a new divisor, and the preceding divisor A as a
new dividend, and divide till a remainder is again obtained
of lower dimensions than the divisor ; and so on. The last
divisor is the G.C.M.
Before giving the general theory of the G.C.M. we shall
work out a few examples.
Ex. 1. Find the G.C.M. of a? - 6a5 - 27 and 2ar^- lloj- 63.
According to the above rule, the operation is as follows: —
ar» - 6a; - 27)2 aj^ - 11a; -63(2
2ay^- 12a;- 54
a; - 9)a^ - ex- 27(a; + 3
ar' - 9a;
3a;- 27
3 a;- 27
196 ALGEBRA.
Ex. 2. Find the G.C.M. of 10 a? + 31 ar» - 63aj and Ua*
■+ 51a? - 54: X,
"We may tell by inspection that a; is a common factor,
•which we therefore strike out of both, only taking care to
reserve it. The quantities then become —
10a? + 31a; - 63, and 14a? + 51a; - 54.
"We may now proceed according to rule, taking the
former as divisor. We see, however, that the coefficient
of the first term of the dividend is not exactly divisible by
the coefficient of the first term of the divisor. Multiply
therefore (to avoid fractions) the dividend by such a number
as will make it so divisible, viz., by 5. This will not affect
the G.C.M., as 6 is not a factor of the first expression,
viz., 10 a? + 31 a; - 63.
It may as well be here mentioned that the G.C.M.
of two quantities cannot be affected by the multipli-
«cation or division of on/e of the quantities by any
quantity which is not a measure of the other. We
shall, for a similar reason, reject certain factors or introduce
them into any of the remainders or dividends during the
operation, -(See Art. 47).
14a? + 51a; - 54
5
10 a? + 31a; - 63)70 a? + 256 a; - 270(7
70a? + 217a; - 441
38a; + 171
Bejecting tiie factor 19 of this remainder, we have —
2a; + 9)10a? + 31a; - 63(5a; - 7
10a? + 45 a;
- 14a; - 63
- 14a; - 63
Hence, 2 a; + 9 is the last divisor, and multiplying this by
a;, the common measure struck out at the commencement, we
^d the aC,Mtobex(^^x -v^^Qt*l^-v 9a^
GREATEST COMMON MEASURE.
197
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198 ALGEBRA.
Dividing tMs remainder by 14371, and taking the quotient
for a new divisor, we have —
x" - 7 X - 3) " 72a? + 679aP - 1009x •- 525(-72a5 + 175
^ -72a^ + 504a:'+ 216a;
175ai«- 1225 a;- 525
1753;^- 1225 a;- 525
.-. ar* - 7 a; - 3 is the G.C.M.
It will be seen that we have introduced and rejected factors
during the operation in order to avoid fractional coefficients.
This, as will be seen from the general theory, will not affect
the result, provided that no factor thus introduced or rejected
is a measure of the corresponding divisor or dividend, as the
case may be.
Theory of the Greatest Common Measure.
46. Let A and JB be the two algebraical quantities, and the
operation as indicated by the rule (Art. 45) be performed.
Thus, let A be divided by B, with B)A(p
quotient p and remainder G. Then pB
let B be divided by G, with quotient y, G)B(q
and remainder D. Lastly, let G be qC
divided by £>, with quotient r, and D)G(r
remainder zero. '^j^
Then we are required to show that -jr
J9istheG.aM. ofilandA ^
(1.) i) is a common measure of A and B.
Now, we have G = rJD, B = qG + J), A = pB + C.
Hence, 2> is a measure of (7, and therefore of qG. It is
therefore a measure of qG + JD or JB. Hence, also, J9 is a
measure ofpB, and since it is also a measure of G, it must be
a measure of pB + G or A» But we have shown it to be a
measure of B. Hence, 2> is a common measure of A and B,
(2.) 2> is the G.C.M. of u4 and ^;
For every measure of A and B will divide A - pB or C;
and hence every measure of A and B will divide B — qG or
£>. Now, I) cannot be divided by any quantity higher than
J?, and, therefore, there cannot exist a measure of A and B
higher than !)• Hence, D ia^<&Q(«^^«^t Aoad j&.
THEOBT OF THE GREATEST COMMON MEASUBE. 199
47. A fa4it<yr whicJh does not ccmtain any /(wtor com/nwn to
both A (md B may he rejected at any stage of the process.
Let the operation stajid thus: —
B = 7nB ' suppose,
B)A{p C)B!{q
G = n(f suppose, D)G\r
-Q
where neither m nor n contains any factov common to A
and B.
It will be an exercise for the student to show that D is the
G.C.M* of -4 and ^*
48. A f alitor y whicJi luis n>o factor thai tlie divisor futSy may
he introduced into the dividend at any stage of tlie process*
The operation may stand thus — -
B)mA{py where m has no factoi' that B has ;
pB
C)nB(qi where 7i has no factor that C has ;
qG_
D)C{r
rD^
As in Arts; 46, 47, it may he Easily shown that D is thel
G.C.M.
Both the iftbove principles are made use of in working out
Ex. 3 Art 45.
49, When ia, common factor can be found by inspection, it
is advisable td strike it out of the given expressions. Then,
having found by the ordinary process the G.C.M. of the re-
sulting quantities, we must multiply the G.C.M. so found by
the rejected factor.
Thus, 4 a; is (common to the quantities 4 a* — 20 ai* + 24 ai,
and4a^ + 16(b* - 84aj. .
Bejecting it, we get a:* - 6 a; + 6, and ai" + 4«j - 21,
whose G.C.M. is easUy found to be a; - 2.
Midtiplying by 4 a:, we find the reqmred Ci.CULi \ft \s4
4a? - Sxi
200 ALGEBRA«
60. By a little ingenuity on the part of the student in
breaking up the given expressions into factors, the ordinary
and often tedious process of finding the G.C.M. may be
avoided. The limitis of our space will allow us only one
example.
Ex. Find the G.C.M. of 3 a:' + 4ar» - 10 a; + 3, and
15a» + 4:7a^ + 13a;- 12.
The first expression contains a; - 1 as a factor (Art. 30),
for the sum of its coefficients is zero. The other factor may
be obtained thus —
3a:" + 4a;»-10a; + 3 = 3a^-3a^+ 7a;*- 7 x - 3a; + 3
= 3ar^(a;- 1) + 7a;(a; - 1) - 3(a;- 1)
= {dx" + 7a;- 3) (a; - 1).
Now, 3ar'+ 7a; — 3is not further resolvable, and a; - 1
is evidently (Art 30) not a factor of 15 a;* + 34 a^ + 13 a;-
12. It is, therefore, very probable that 3a;' + 7a;-3is the
G.C.M. required.
We may test it thus —
15a;» + 47ar^+ 13a;- 12 = 15a;' + SSar' - 15a;+ 12ar» + 28a;- 12-
= 5a;(3a;" + 7a;-3) + 4(3a;' + 7a;-3)
= (5a; + 4)(3a;«+ 7a;- 3).
Hence, 3a:*+ 7a;- 3is the G.C.M. required.
G.C.M. of Three or More Quantities.
51. The G.C.M. of three or more quantities may be foimd
thus —
Rule. — Find the G.C.M. of any two of the quantities, then
the G.C.M. of the G.C.M. so found and a third quantity, and
so on. The last found G.C.M. will be the G.C.M. required.
Ex. XII.
Find the G.C.M. of the following—
1. ar* - 5 a; + 6 and ar* + 3 a; — 18.
2. a;' + 6a;»+lla; + 6anda;' + 5ar* + 7a; + 3.
3. 2 3;^+ lOar*- 18a;-90and3a;»+16ai»-26a; - 141.
4. a^ + {a + ba: + ab and aP -{- (a ■{- c) x ■{- ac,
6. a? - l^ and a' + a?b -v ab-.
LEAST COMMON MULTIPLE. 201
6. a* - 4 « + 3 and a' + 4 ar* - 5.
7. 4a;»-. Z2a^ + 85 aj - 75 a^nd 3 ic* - 15 ar» + 16a; + 9.
8. 9ar» - 3a^ + 2y- 4and6a;* - ia^ - 9xtf + 6y'.
9. 48 aJ* + 8 aj' + 31 ar» + 15 a: and 24 a;* + 22 aj^ + 17 a;'
+ 5 a?.
10. 15 a» + a^S - 3 aJ^ + 2 6^ and 54a^&» - 24 5*.
11. 3ar»-(3c + ci+l)ar'-(2a + 5-3c-c? + 2)a;
+ 2a + h and 2ar*-(a+6 + 2)a: + a + 6.
12. 6a^ - 4aj* - 11 a;^ - 3ar^ - 3aj - land4aj* + 2a?
- 18 ar» + 3 a: - 5.
13. a6 + 2 a2 - 3 6^ _ 4 5c - oc - cr' and 9 ac + 2 a^ -
5a6 + 4c' + 8 6c - 12 6^.
14. 6*af* + e* + a^ + 1 and c^^a;* - 6^ + a;* - 1.
15. aa^ + (6+c)ar*-aaj-5 — c and ea? - {f - g) a? +
(/ - e) a; - ^.
16. 4 aj* + 2 a^^ + 4 ar* + 39 a: - 9, 8 a;* + 20 a:' + 51 03
+ 9, and 2a^ + a^+3ai'+18a?.
17. aa? - (c + 1) aj' + (c + 1) a; - a, 5a* - (6 + c?) ar* +
(c + d^ ai" — (c + c) a; + c, and (c + 1) a^ + (c? + 2) ai* -
ld+ 1) a? - (c + 2) ar».
18. a^ - 6^ + c* + 3 a6c and a^ - 5* + c' + 2 oc.
Least Common Multiple.
52. When two or more algebraical expressions are arranged
according to the^jpowers of some letter, the expression of
lowest dimensions which is divisible by each of the given
expressions is called the L.C.M.
53. The L.C.M. of monomials and of expressions whose
factors are apparent may be found by inspection.
Ex. 1. Find the L.C.M. of ab, ac, ad, be, bd, cd.
If we form an expression, whose elementary factors con-
tain each of the elementary factors of the given quantities,
we shall evidently have a common multiple ; and if no ele*
mentary factor of this expression is of a higher power than
the highest power of the same factor in the given quantities.
we shaQ get the L.C.M.
Sence^ the required L.C.M. = abcd.
202 ALQEBRA.
Ex. 2. Find the L.C.M. of—
(a - 6) (6 - c), (a - 5) (c — a), (6 - c) (c - a).
Ans. (a - 5) (6 — c) (c - a).
Ex. 3. Find the KCM. of a {x + 1), h {a? - 1),
c (jc* + 2 a; — 3), c? (a;* + 4 a; - 3). We may write the given
expressions thus —
a (a; + 1), 5 (aj + 1) {x - 1),
c (a: - 1) (a; + 3), c? (a; + 1) (a; + 3).
Hence, the reqnii'ed L.C.M. = abed (a; - 1) (aj + 1)
(a; + 3).
Ex. 4. Find the L.C.M. of a' - oa; + a?, a' + aa; + ar*,
a* + a^, a' - a^.
Now (Art. 29) a^ + a* = ^a + a) ^a' - oa; + a:*),
and a' - a;* = (a - a;) (a* + aa; + ar*).
Hence the required L.C.M. —
:= (a + a?) (a - a;) (c? + aa; + a:^) (a? - aaJ + a?) = a* - a:^.
64. j?%d L.CM. qf two quantities is /oimd hy dividing
their product by their G,G,Mi
Let a and h be the two quantities, and d the G.C.M. ;
And suppose a = pd and b = g^c?.
It is evident that p and ^ contain no common factor.
Sence pq is the L.C.M. of p and q ; and, therefore, no expres-
sion of lower dimensions ttian pqd can possibly be divisible
by jpd and qd.
Hence pqd is the L.C.M. otpd and qd, or of a and 6.
l^ow pqd = pd X g(Z-fc2=ax6-rC?, and hence the rule;
56. To find the L.C.M. of three or more quantities.
• Rule.— Find the L.C.M. bf two of the quantities, theii
the L.C!.M. of thei expression thus obtained and a third
quantity, and so on. The last expression so found is the
L.C.M. required.
We fehall prove this rule in the case of three quantities.
Let a, by c he the quantities, and m be the L.C.M. of a
and b.
Then the L.C.M. of m and e is the L.C.M. required.
For every common multiple of m and c is a common
iiiultipie of a, 6, c. And evexy commQrLT£wi^\^^^io.wxd6
LEAST COMMON MULTIPLE. 203
must contain the m, their least common multiple. Hence,
every conunon multiple of a, b, c must be a common multiple
of w and c, and the converse is also true. Hence, the
L.C.M. of m and c is the L.C.M. of a, 5, c.
Ex. xni.
Find the L.C.M. of— -
2. 5 d?h\ 6 aV, 4 6V.
3. (a - 5) (6 - c), (6 - a) (a - c), (c - a) (c - 6).
i, aai(x •{- a), a'(a5 - a), cc* - a^
5. a'+3a; + 2, a;' + 4a; + 3, a? + 5a; + 6.
6. 0? - a; - 30, a:" - lloj + 30, a? - 25.
7. Gai" + 37 aj + 66, 8ar» + 38aj + 35, 12ai» + 47a;
+ 40.
8. 5(ar» - a; + 1), 6(0* + 1), 7(aj^ + 1).
9» aJ* + aV + a\ aW + a^x ^ a\aa? - a^x + a*.
10. ar* + (a + h) x + 06, o^ + (c» + c) a; + oc, ar* +
(6 + c)a; + 5c.
11. 1 - a?, 1 + a?, 1 + ar^, 1 + a*, 1 + a?.
12. «»+ 6a» + 11a; + 6, ar» - 6ar» - 25a; + 150*
13. a» - 3a5 (a - 6) - 6', a'* - 6', aJ^ + a'6 + aft'*.
14. a;* - 1, Ga;* + 5a;* + 8a;» + 4ar* + 2a; - 1.
15. a* - 2a'6» + 5*, a* + 4a'6 + Ga^ft^ + 4a5» + 6*,
a* - 4a»6 + 6a^6* - 4a5' + 6*.
16. 3a;> - 4a; + 1, 23;^ - 7a; + 5, 4a;* + G a;' +
lOa;.
17. 3ar» + 6 a;- 24, a;* - 12 a; ^ 16j 5 a;* - 22 a; - 36.
18. a' - aW, U" - a\ ah'' - 6»^ a^ft - a».
19. 3a;* - 48, ho? - 20, 3a;» - 16a; + 20.
20. a;* - 2/*, a;* - 2 o^'if + «/*> »* + a;^y + a^ + 2/^.
21. a;" + CKC* + aV + aW + a*a; + a' and a;* - aa;^ +
aV - aW + a*a; - a';
22. a' + 52 _ c« - d« + 2a6 - 2ccJ and a* - tf - *
204 ALGEBRA.
23. a' + 6« + c" - 3 abc and {a + 5)^ + 2 (a + 6)o
+ c*.
24. (a + hf - (c + c£)«, (a + cf - (6 - ci)», (a + df
- (6 - c)^ (c + df - (a - 6)', (6 + df - (a - c)», and
(6 + cy - (a - c?)«.
CHAPTER V.
Fractions.
56. It is mmecessary to repeat here the propositions relat-
ing to fractions which were proved in Ariflimetic, Chap. II.
of this work. The student will see that, by substituting
general symbols for the particular figures there used, the
reasoning will equally hold. We shall work out a few
examples to show the method of dealing with them in algebra.
a' - 2»" + aj — 12
Ex. 1. Simplify the fraction — o o — ZTTS — '
By inspection (Art. 30) we see that a; - 3 is a factor of
numerator and denominator. We have then —
a? - 2a^ + a;-12 ^ a?{x - 3) + a; (a; - 3) -f 4 (a; - 3 )
ar^ + 2a; - 15 (a: - 3) (a; + 5)
_ (a^ + a? + 4) (a; — 3) _ ar* + a? + 4
(a; - 3) (a; + 6) '^TTb -^^•
Ex. 2. Fmd the value of — — r- +
a + b a '- h d^ + b^'
1 1 2a __ {a " b) + (a + b) ^ 2a
a + b a ^ b a^ + b^" {a + b) (a - b) a* + U^
= 2a 2a ^ / 1 1 \
a« - 6^ "^ a» + 6« "• ^ " \,a' - 6^ ■" a* + by
(a^ + b^ - (a° - b") 2V iab^
- ^^' (a^ - b^) {^a^ + b^^ "" ^^' a* - ^* "" a* - b''
FRACTIONS. 205
be CLC
Ex. 3. Find the value of , tw \ + ti wi 1
{a - 0) {a — c) (6 - a) (o - c)
ah
"^ {c-a) {c- b)'
The second denominator has a factor, (6 - a), which differs
from a factor, (a - 5), of the first denominator in sign only.
We shall therefore change the sign of the second fraction,
and also of its first factor. This will not alter its value.
And, similarly, we find that by changing the signs of each
of the fEU^rs of the third denominator we shall have them in
a form corresponding to factors of the first and second denomi-
nators. The sign of the third fraction will not be changed, as
the sign of the denominator will, on the whole, be unchanged.
The given expression then will stand thus —
be ac ab
" (S":^)~(a - c) " (a - 6) (6 - c) "*" JaT^c) (6 - c)
be (b - c) " ac {a - c) + (ib(a - b)
" {a - b) (a " c) (6 - c)
be (b — c) " a^c + ac^ + a^b - ab^
= (« -b){a- c) (6 - c) ' "'•' w-««™i»g»ng.
a^(b - c) - a (b^ - e^ + be(b - c) ., ,. . ,.
= — ^^ — , rv / TTi r -9 *"®^> dividing nume*
{a - b) {a - c) {b - c) °
rator and denominator by 6 - c,
^ a^ '- a{b + c) + be _ {a - b) (a - c)
" ~{^b)~(cr^j~ ~ (a - 6) (a - c) = ^'
Ex. 4, Simplify —
f J _ iaJ'x " Sax" + 0^ I x / a' + ^«'^ +300:^ + a^ )
The given expression —
a'(« + x)'-' 4a^x+ Sao?- a? fl^(a - aj) + 4 a'jc + 3 oo^ + a'
a + X a — X
_ a' -i; a'a; — 4 a^a; + 3 oaj* - jc* a^ ■- a^x + 4: a^x + 3 aa^ + a^
" a i- X tt - X
206 ALGEBRA.
a^ - 3a^x + 3ax^ - a^ a^ + 3 a^a; + 3 ooj? + as?
2-. X
a + X a - X
a + X a ^ X 1 1 ^ '*
Ex. 5. Divide —
Now —
/ z.\ ( ^ , ^ \ / 7x a(a5 + 6)^ + b(x - of
or, reducing —
(a; - ay {x + by (a? - a) (a? + b)
_ (a + b) {x " a) (x + b)
{a - 6) (ar* + a6)
Ex. XIV.
Simplify the following expressions: —
,a5^-5a; + 4 a'-3aj + 2
* x^ + 2x - 2i'a^ + 4ar^- 5*
g-^ Gar* + 29a; + 35 2ar^ + 7a; - 9
* 14ar^ + 39a; + 10' 6a;' - 3ar^ - 4a; + 2'
3 g* - 2a^P + b^ 24a^ ~ 28a^6 + 6a&° - 75>
; a^ - 4a26 + 4a62 - b^' 6 a^ + 11 a6 - 21 6^
* ar»(y + «)5~a^(23/*+33/« + ^) +2^(y + »)' a;* - 2/^ *
;; 1 ^ 1 r 1
+ b a-oa-o a-vo
FRACTIONS. 207
/•aft ab ab
' a + b a - b ab - b^ a^ + ab'
/- 1_ X - I ^ X " I
' 4^^T) 4 (ar^ + 1) 2 {x" + if
g _5 2__ 18 _ 7x + 1
* a; + 1 {x + 2f ' 5{x + 2) b (x" + I)
9 1^ 11 _ U
' x + I " X + 3 " {x + If
.Q _8 4 2 16a; -f 14 _ lGa;-8
' (a;-l)* (a;-l)3 ^^^3+3(^_l) 3(^-a;+l)-
11. 1 + 1 -, __!___.
(a - 6) (a - c) (6 - a) (6 - c) (c - a) (c - 6)
12. ^ + ^ + <^
(a - 6) (a - c) (6 - a) (6 - c) (c - a) (c - 6)'
ft2 ^3 r^
13. ^ + ^ + ^
(a - 6) (a — c) (6 - a) (6 - c) (c - a) (c - ft)*
14.
a^ ft3 c»
(a - ft) (a - c) (ft - a) (ft - c) (c - a) (c - ft)*
jg (a + iy + (b-cY + {a + cY _ 2 _ 2 ^ 2 '
(a + ft) (ft - c) (a + c) a + c ft - c a + b*
17- {— ■^— +-^^M-^^ — ^-4^.}.
{a + x a-x a^ + ar ) {.a-^-x a-x a^ + xr )
{ a^ "' (r a + ftj I a^ - ¥ a + ftj
19 ' / 1 -?^ + 2^* I / 2a^ _^ 2a^ _^ , )
I a* aV + ar')/ "" I ar^(a2-ar^) v? ]'
a»(^-+l)+ftY-+l) + c^(^f +l)
on « , ft , <5 ^c / \a / Vft • /
c a ft aft + oc + CO,
208 ALGEBRA.
"•{c-^^)'t'}-{(^:)'->}-{(^D'*'}
00 /1-a^ ■ l-x \ ^ / 1+a; l-a^}
23.
(a + &)(» + c) (x -a) (a+ b) (6 "' c) {x + h)
y.3
(a + c) {b — c) (x + c)
(a- a s(r ) { cir a x )
27. ^' + ^' ^ 2& _ 2a ^g
' (a + 6)^ {a - by a + b a - b
28.
5* ^ .7^ - 2ay ^ j^
(a; - y) (a + x) (x ^ y) (a + y) (a + ?/)'
00 y ^ 2/ _5 (^ ^\
2^ ?~
'•C-7TV}(-^i>
30.
y + —
SIMPLE EQUATIONS. 209
CHAPTER VI.
Simple Equations.
57. When two algebraical expressions are connected by the
sign ( = ), they are said to form an equation.
When the equality is such that it is true for all values of
the letters in the given expressions, it is called an identity.
Thus, {x + a){x + h)=a?-¥{a + h)x + ab \ ij^j^tuies
and {a + hf - (a - hf = 4.ah | are laentities.
58. When the condition of equality is such that some one
or other of the letters must have paHicular values or a
Umited niunber of values, thQ statement of equality is termed
an equation of condition, or, more briefly, an eqtuition»
Thus, it may be found on trial that the equality
4:X + 2 = 3x + 5
is true only when oj = 3. Such an expression is therefore
an equation.
69. The letters of an equation to which particular or a
limited number of values must be given are termed unlcnoivn
quantities.
Equations may contain one, fwo, three, or more unknown
quantitie&
The determination of the particular value or values of the
unknown quantities is called the solution of the equation,
and each of the values which satisfies the equation is said to
be a root of the equation.
60. The expressions on the left and right sides of the sign
( = ) are termed the first and second sides respectively. It
follows, therefore, that —
1. If both sides of an equation he multiplied by the same
quantity, the equation stiU subsists.
2. If both sides be divided by the same quantity, the equor
tion StiU holds,
3. Any term may be l/i^ansposed from one side to tlie other
if the sign of the term be changed.
Thus, if 3 cc + a = 5, we must have also
Zx = b - a,
5 ■
210 ALGEBBA.
for this results from subtracting a from each side of the
equation.
4. The equation lioMa if every term on both sides has its
sign changed.
Thus, if ooj + 6 = CO? - cf, we may reason as follows: —
The quantity (ax + b) looked upon as a whole is given
equal to the quantity (ex — d) looked upon as a whole. If
we change the qualities of these quantities, they will evidently
be stUl equal.
Hence, - (ax + 5) = — (ex - d)
or, - aa; - 6 = — ex + d,
Now, this is the result of changing the sign of every term on
both sides of the given equation.
5. The sides of cm eqwxtion may be reversed vnthmU destroy-
ing the eqv^dity.
Thus, if mx + n = px ■{■ q^ii must also follow that
px -v q = mx + n,
6. TJie sides of an equation may be raised to the same
POWEB, or we may extract the same root of both sides, arid
the equation still subsists.
61. Simple equations are those in which the unknown
quantities are not higher than the first degree, when the
equations are reduced to a rational integral form.
The following is the general method adopted in solving a
simple equation involving only one unknown quantity —
1. Clear qffra>ctions if necessary.
2. Transpose aU the terms involving the unknoum quantity
to the first side of the eqiuition, and all the remainiruf term
to the second side,
3. Simplify both sides if necessary , and divide both sides
by the coefficient of the vnknotvn quantity,
Ex. 1. Solve the equation 5aj+6 = 3aj+12.
Transposing the terms, we have—
6a;-3aj=12-6.
Now, simplifying, we get —
2x = 6;
and dividing each side by the coefficient of the unknown
quantity, viz., by 2, we have —
X = ^ ^ ^ = ^.
SIMPLE EQUATIONS. 211
Verification. — Putting the value 3 for a; in each side of the
given equation, the firstside becomes 6x3 + 6 or 21; and
the second side becomes 3 x 3 + 12 or 21. The value of aj
found therefore satisfies the given equation.
Ex 2. Given ^^-^ h- | = 20 - ^-^, find x.
Clearing of fiuctions, by multiplying every term on each
side by the L.C.M. of the denominators, viz., by 6, we get —
3 (x - 2) + 2 a; = 20 X 6 - 3 (aj - 6).
(Beginners often neglect to multiply integral terms such as 20 by
the L.C.M.)
or3aj— 6+2a;= 120 - 3 a; + 18, or, transposing,
3 a: + 2 a; + 3a;*= 120 + 18 + 6, or, simplifying,
8'a; = 144,
or dividing each side by 8, tha coefficient of x, we have —
aj = 18, the value required.
(It will be good practice for the student to verify this result as in
the last example).
„ «4aj-21 ^ Ta;-28 „, 9-7aj
Ex. 3. ^ — + 7-J + g — = a? + 3f g — + TV-
It is sometimes convenient to first pcurtially clear off fi*ac-
tiona Thus, multiplying each side by 72, we have —
72 (4aj- 21)
-^^ '- + 47 X 12 + 24 (7 a; - 28)
= 72a; + 15 X 18 - 9(9 - 7a;) + 6;
2880;
or ^^ - 72 X 3 + 564 + 168 a; - 672,
= 72 a; + 270 - 81 + 63 a; + 6;
or, transposing,
41+ aj + 168 a; - 72 a; - 63 a;
= 270 - 81 + 6 + 216 - 564 + 672 ;
or, simplifying,
74^1- X = 519 ; or, multiplying each side l)y 7,
519 a; = 519 x 7;
.\af = 7.
212 ALGEBRA.
Ex. XIV. ^
1. 5a; + 2 = 2a; + 11.
^- 4 "*■ 6 "^ 8 " '^^•
3. 2a; + a=3a;-6.
4. 3(aj-7) + 4a;=2(2aj- 4) + 2.
a; - 1 a; + 3 _ 2 a; - 7 8 a; - 1
^' 2 ■*" 4 - 6 "*■ 12 •
Ta; - 8 3 a; - 2 _ 10 a; + 3 7 a; - 3
^- 13 "*■ 7 " 11 " 18 •
4 a; - 15 2 a; -f 3 _ 5 a; - 1
7. g Q - 12 " ^**
8 (5 a; + 2) 2 a; - 1 _ 17 a; - 2 5 | -f 80 a ;
^' 3 ""8~4"*" 7'
9. ax + be = hx + ac,
^ X a X h
lO. - + ^ = V + -.
abba
,, a;-a x - h ^ ex ^ <?
12. afta; + 5» = 6'a; +a».
a; a; a; . ,
13. - + T + - = «^ + «c + 6c.
_ . a; + 5 rt + a;
14.
a
h
, ^ aa; + fta; + ca; .
15. 1 = a + 6 + c.
^' a? - Zhx _, 5a; 65a;-5a' 5a; + 4fl
16. a; - 2 — - 6' = — - + jr- « — " " — i •
a^ a 2 a* 4 a
17. -15 a; + -025 = -075 3; + -175.
,^ 3 - -125 X -V 2 + '1875 x ^ •
18. g \ -IQ - ^ -083,
ABBREVIATED METHODS FOR PARTICULAR CASES. 213
20. (x + a) (x + b) = (x + c) (x + d).
21. (x'-a) (x-h) = (x - cTThy.
oo; oa5 co; * ^ ^ •* ^ocos oca; mx/
^^ X - a X - b aj-c'^o/l 1 1\
oc oc ao \a c/
^, I - ax I - bx I - ex /2 2 2\
24. — 7 — + + — r— = ( - + r + - y^-
6c ac ao \a b c^
Abbreviated Methods for Particular Cases.
62. When the unknown quantity is involved in both
mtmerator and denominator, it is often convenient to reduce
such fractions to mixed numbers.
6a; - 7 12 a; + 18
Ex. Solve the equation — —fy' S x - 5 ^
15 J- • • 6 a; - 7 . ^ 19
By division we get -^^-:^2 " ^ " ^TI^
12 a; + 18 , 38
and -s K- = 4 +
3a; - 5 "" ' ^ 3a;-5*
Hence the given equation becomes —
19 . 38 „
or 6 - — r-o - 4 - o"- — p = 2 ;
x + 2 ox - 5 '
19 38 • , ,
or, transposing, - ^-^ = S^ITg + 2-6 + 4;
19 _ 38 .
^^ " a; + 2 '^ 3 a; - 5'
or, dividing each side by - 19, we have —
1 2
a: h 2 '^ ." 3x - b'
214 ALGEBRA.
Hence, multiplying each side by the L.C.M. of the de-
nominators—
3a;-5=-2(a; + 2)=-2a;-4,
or 3 a; + 2 a: = — 4 + 5,
or 5 a; = 1,
63. When each side of an equation consists solely of a single
fraction, the nwmerator of either fraction m/ay change places
with the denominator of the oiheir, *
Let ^ •= ~ be the equation.
Multiply each side by 6, then, by Ari 60 (1.) —
a p pb
. b q ' q
Divide each side by p, then, by Art. 60 (2.) —
a pb a b
- = — -r 7?, or - = -
p q ^' p q
Here the denominator b of the first side of the given equation
has changed places with the numerator p of the second side.
q p
And similarly we may show that I = - , where the other
6 a
numerator and denominator have changed places.
Cor. Tlie two sides of an equation of the form i = ~ fnay
be inverted.
For interchanging « and b in the last result, viz., f = --,
a
n h
we get - = -, and therefore also, by Art, 60 (5.), we have
^ = ?
a p
(The student is cautioned affainst inverting the separate terms of
the two sides of an equation when there are more than one term on
each side. )
64. When each side of an equation consists solely of a
single j&action^ we may "perfoxai ^'ft io\krwix\i^ operations : —
ABBREVIATED METHODS FOB PARTICULAB CASES. 215
1. We may add or avhi/raot the numerator arid denomiruxtor
of 'RA.ca fraction for a new numerator or denominator, and
retain either the wiginal nwnwrator or denominator for the
other term of the fraction, both sides being always similarly
treated.
Thus, if T = -, we have —
(O -5—= —^y (^)— r~ == ~7"'
(m.) = , (IV.) = ,
or, (v.) we may have equations formed by inverting each of
these.
These results are easily obtained —
For, since t = "^, we have, adding unity to each side —
a ^ p , a + b P + g
7 + 1 = - + 1 or — J—- = -,
b q ^ g
And so, by subtracting unity from each side, we get —
a ^ b p - q
— I — = , and so on.
b q '
2. We may take the sums of the numerator and denominator
of each for nefw numerators or denominators, a/nd the dif-
ferences ybr the oifier terms of the fraction; and vice versa,
both sides being always similarly treated.
Thus, if? = ^, we have also ^* = '^,
' b d* a -- b c - d
J a - b c - d
and — - — r = — ;; — >>
a + 6 c -^ d
^ « ..t 1.^ + ^ p + q
for we have just shown that — r — = >
, a - 6 p - q
and — J — = -■ •
b q
216 ALGEBRA.
Hence, dividing equals by equals, we get —
a + b ^ a - h p + q ^ p - q
h ' ^ ^ ^ ' 9 ^
or r = -, wHcli is the first result.
a - P - q
And inverting each side, we have, by Ari 63 (Cor.)—
a - b p - q
a + b " p + q
^ ^«,., .. mx + a + b nix + a + e
Ex. 1. Solve the equation 3 = r y
^ nx - c - a nx — b - a
_^ , ^^ , mx + a + b nx - c - d
By Art 63, we have = f r
•^ ' mx + a + c nx — - a
Then, by Art. 64 (1.), retaining the numerators and taking
the differences for new denominators, we have —
Tiix + a -^ b nx - c - d
b " c b - c ^
or, multiplying each side by (6 - c) —
mx + a + 6 = 7ZX - c - cZ; or, transposing —
/ \ /r n a + b + c + d
(m - n) X = - (a + b + c + a): .-. a; =
Ex. 2. Solve - c_^rr=^.:_ r— = a.
ija + a; - ^/a - x
We may consider the quantity a as a fraction whose deno-
minator is unity y or as - .
Then, Art. 64 (2.), taking the sum and difference,yfe have—
2 Ja + X a +. 1
— - = .. or —
^Ja - X a - V
hja + 03 • a + \
—, — = i ; or, squarmg-—
a + X _ a^ + 2a + 1
a — « a?-^a-vV
ABfiHEVlATED METfiODS fon PARTlOULAft CASES. 217
Again, taking the difference and sum, we have—
205 4a
— = ; or —
2a 2«» + 2'
X 2a ' ''
a a^ + 1
2a^
,, X — — s = •
a^ + 1
65. We now give an example to show that sometimes the
easy solution depends on an advantageous arrangement of
the terms on the two sides of the equation.
Ex. Solve Jx + 4: + Jx~^^ = 7.
Transposing, we have —
^x + 4 = 7 - fjx - 3 ; squaring, then —
a; + 4 = 49 - 14 Jx~^^ + (a; - 3);
subtracting x from each side and transposing, then —
14 Jx"^^ = 49 - 3 - 4 = 42.
.'. tjx - 3 = 3 ; or squaring,
a; - 3 = 9; .-. a; = 3 + 9 = 12.
Should the student commence by squaring at once, he will
render the equation more complicated.
1.
Ex. XVI.
3aj + 7 3aj - 13
a; + 4 33-4
2. (a; - a) (a; - 6) = (a; - c) {x - d),
3x +13 3a; + 10 x
3.
15 dx - 50 5'
. 1 - 25a; 3 - 2^a; _ 28 - 5a; 10a; - 11 x
15 ^14(a;-l) 3 "* 30 3'
5 ^ " ^ 3a; - 13 ^ 1
' 6x + 5 ISx - 6 3'
6 3 ^ 5 ' 2x _ J __ 4 a;^ - 2
I - 2a; 7 - 2x 7 - I6x -V ^:X^
a*
2l8 ALQ£BtU4
- 18a3 - 22 , ^ .1 + 1605 ^a, lOl - 64a;
^- 13-20: ■" ^^ -*■ —8— = ^^* 8 '
8 ^^ + ^ + 58^ + Ux ^ g
" 3o; + 1 9 + 2o;
9.
3 . U
4o; - 9 60; - 21 X " 2^*
.r. 4o; - 7 2 - Ux Sj + a __ 10 - 3^o; 19
2o; - 9 7 14 2 21
11 ^^ - 7a; - 6| _ 90;^ - 12a; - 19 ^ 17
"2a; - 3 3o; - 6 6a; - 7'
To«a; + m+l ax + n (tx + m
iz. + — - = +
ax + m - I ax + n - 2 ax + m - 2
ax + n + I
oa; + 7* — 1*
^/x - a + 6 - s/x + a - b a — h.
13.
Jx - a + b + *Jx + a " b a + b
1 - Vl - Vl - a; ^
14. / v^-zE ZF = 6.
1 + V 1 - n/1 - a;
15. J2x + 10 + N/2a; - 2 = 6.
3
16. ^S - X " J-, -_--. = ijl - X.
17. 4TT~^ + 4^1 - V«^ = 2.
,^ aa; + 1 + VaV - 1 ,
18. ^ r, , , = ^•
aa; + 1 — va ar - 1
a II h _ a - h
^^* VaJ - Jb "^ Va; - Ja " "jx"
20. —1 7 -p TT + Va.
six + xja \fx - v^
//s/a; + a _ / ^/a; - a 1 -
PROBLEMS INVOLVING ONE UNKNOWN QXTAi^flTY. 210
"-" a - 3 " .6 - 5a; + ar» ~ *^ ~ a; - 3*
4a; + 5 a; + 5 2a; + 5 a;^ - 10
a;+l a; + 4 x ■¥ 2 a; + 3
^,005 + 6 ca; + e a^ + c*
ex + d ax +j ac
Problems producing Simple Equations involving One
Unknown Quantity.
66. To solve an algebraical problem we represent the re-
quired or unknown quantity by a letter, as x, and then ex-
press the given conditions in algebraical language. Thus wo
form an equation, the solution of which gives the required
value of the unknown quantity.
Ex. 1. My purse and money are together worth 24 shillings,
and the money is worth seven times the purse. Find the
value of each.
Let X = the value in shillings of the purse,
Then 7x = „ „ money.
Now, by problem tiie value of both together is 24 shillings.
Hence we have —
a; + 7a = 24
or 8a; = 24
.*. a; = 3, the value in shillings of the purse,
and .*. also 7 a; = 7 x 3 = 21, „ money.
Ex. 2. What number is that to which, if 36 be added, the
sum shall be equal to 3 times the number ?
Let X = the number ;
.\ X + 36 = the sum when 36 is added,
and 3 a; = 3 times the number.
Hence, by problem —
a; + 36 = 3 a;,
or a? - 3 a; = - 36,
or - 2 a; = - 36;
— *\(\
.«. a; = ^ — - = 18, the number rec\\\iYcd.
Ex. 3. Ik 4ir^smTk0^ between tvo tofwns is sucii that a train,
viiose e^oA is 30 izuks an boar, takes 1 hour nuve in going
10 miles otct 5 times the disranop than a train whose speed
is 20 miles an hour takes in going within 4 miles of 3 tunes
the distanceL Find the dif^amy between the town&
]>t X = the diwancp required in miles.
Then 5 jr -^ 10 = 5 times the distainfie together with 10 miles,
o X -^ 10
jind — - = time in bonis to travel this distance at 30
miles an boor.
AndsOy — -- — = time in bonis to travel 4 miles less
than 3 times the reqniied distance, at 20 miles an hour.
Bnt by the problem the former of these times exceed the
latter by 1 hour.
Hence^^l«-^?^=1.
30 20
From this we easily find a; = 28. .
Hence 28 miles is the distance required.
Ex. 4. Find the price of an article, when as many can be
bought for Is. 4d. as can be bought for 2s. after the price
has been raised Id.
Let X = the price required in pence;
Then -- = number of articles bought for Is. 4d.
And X + \ = the raised price in joerice.
24
/. j^ = number of articles bought for 2s. at the raised
a + 1
price*
But, by the problem, the number of articles in each case is
tho same*
Ilenco — = -"^ — , from which aj = 2.
(0 a; + 1
Henco 2d. is the price required.
Ex. 5. A man sells geese at as many shillings each as the
number he has, and liavmg t^V.\xmfi^ ^^^ ^da Qiat if he had
-■/■■
PROBLEMS INVOLVING ONE UNKNOWN QUANTITY. 221
had 2 more to sell on the same condition, and had returned
38., he would have had 38s. more. How many had he]
Let X = the number required;
Then a* - 5 = number of shillings received.
Also, on the second supposition,
(a? + 2)' - 3 = number of shillings he would have received.
Now, by the problem, this latter number is 38 more than
the former.
Hence (a; + 2)^ - 3 = ar' - 5 + 38, from which we find
a: = 8.
Ex. 6. A waterman finds that he can row 5 miles in |
hour with the tide, and that it takes him 1^ hours to row
the same distance against the tide when it is but half as
strong. What is the velocity of the tide 1
Let X = the velocity of the tide in miles per hour.
Now the velocity of the boat when going with the tide
.'. Y - 35 = velocity of the boat when there is no tide.
Again, velocity of boat against the tide when it is half as
sti-ong = 5 ~ 1^ = Y'
10 X
• ' "q '*" o ~ velocity of the boat, when there is no tide.
Hence we have —
10 . a; 20 ^ ^ , . ,
-_. +-- = _.— a; : from which
3 2 3 '
X = 2J.
.*. The velocity required is 2f miles per hour.
Ex. XVII.
1. If I add 25 to 3 times a certain number, I obtain the
same result as if I subti*act 25 from 8 times the number.
Find the number.
2. Divide 70 into 2 such parts that the one shall be as
much above half the number as the other is above 15.
3. Divide X720 among A, B, and C, so that B m^c^ Wn^
twice aa much m C, and A fts much as B aniV G ^o^jiAOtifcx,
222 ALGEBRA,
4. There are two trains, one of which goes 5 miles an
hour faster than the other, and the former performs a journey
of 100 miles, while the latter goes 75 miles. Find their
respective rates,
5. A horse when let out for hire brings in a clear gain of
10s. per day, but costs Is. 6d. daily for food. At the end of
30 days his master had gained £11. lis. Kequired tbe
number of days for which he was hired.
G. A and B have 4 guineas between them, and play at
hazard. B loses -} of his money, and afterwards gains -^ of
what he then had. It is then seen that B has as much
money as A had at the end of the first game. How much
had each at first?
7. A and B have respectively an equal number of florins
and crowns. B pays a debt of 4s. to A, and then A's money
is just half B's. Find what each had.
8. A workman, instead of adopting the 9 hours' sjnstem,
worked 10 hours daily, and had a corresponding rise of
wages. By this means his wages were increased 4s. weekly.
Find his original wages.
9. A person who has regular wages of 26s. weekly, think-
ing to better himself, takes a job at higher wages. He is,
however, put on half-time during 20 weeks of the year, and
finds himself at the end of the year £4, 12s. worse off. Re-
quired his increased wages.
1 0. A company of men, arranged in a hollow square 4 deep,
numbered 144. What was the number in a side of the
square 1
11. In an examination paper there were two series of
questions, and the questions of the second series carried each
3 marks more than those of the first series. A candidate who
attempted 3 of the first series, obtaining half marks for them,
and 5 of the second series, obtaining for these full marks, got
altogether 80 marks. Find the nimiber of marks attached to
each question of the first series.
12. A grocer has tea at 3s. 4d. and at 4s. He selk
altogether 64 pounds, thereby realizing £12. How much did
be sell of each ]
PROBLEMS INVOLVING ONE UNKNOWN QUANTITY, 223
13. If 11 be subtracted from 5 times a certain number,
and the remainder divided by 6, the quotient will exceed by 2
the quotient obtained by subtracting 3 from 4 times the num-
ber and dividing the remainder by 7. Find the number.
14. A garrison of 1,250 men were provisioned for 64 days;
but after 22 days a certain number were called away, and it
was found that the remaining provisions lasted the number
left for 70 days. Find the number told off.
15. At a railway station £15 was taken for single fares,
and £33. 15s. for returns. The number of return tickets
exceeded the single tickets by 10, and the price of a return
ticket was half as much again as a single ticket. Find the
fiai'e for a single journey.
16. In a tour lately made round the world, the distance
travelled by water was 20,000 miles, and by land 8,000 miles;
and the whole time taken was 220 days. Supposing the rate
by water to be two-thirds of that by land, find the number
of days travelled by land.
17. The distance between A and B is 32 miles. A person
starting from A, at the rate of 4 miles an hour, meets another
who started from B half an hour later, at a rate of 3^ miles
an hour. At what point will they meet?
18. There are two clocks, one of which gains twice as much
per day as the second loses, and they are set right at noon
on Monday. When it is noon on Thursday by the first clock,
it is 11*50 A. M. by the second. What is the gain per day of
the first clock]
19. A draper raises his goods a certain rate per cent., and
afterwards reduces them to the original price by lowering them
13^y per cent. Find the original rise per cent.
20. Bequired the distance between two towns such that a
person can perform the journey one hour sooner when he
walks 4 miles an hour than when he walks 3^ miles an hour?
21. The sum of £12. 15s. is paid away with an equal
number of sovereigns, crowns, and sixpences. Required the
aumber of each.
22. A walks along an inclined plane at a cctt^Lm x^\i&^ ^;xA
B walks aloii^ the Imae of the plane at a rate oi oxife->i5Ka^ ol
224 ALGEBRA.
a mile per hour less tlian A. The iuclination of the plane is
guch that A is always vertically over B, and that at the end
of half an hour they are exactly five-sixths of a mile apart
Find the respective rates of A and B.
23. There is a direct road over a hill between two stations
at the foot of each side. The distance on the one side from
the foot to the top is 5 miles, and the road down the other
side forms a right angle with the road up. It is also known
to be 1 mile less down the hill than the direct distance by
tunnel between the two stations. Find the distance down
the hiU.
24. Two trains, whose respective lengths are 122 and 98
yards, and the former of which is going at the rate of 35 miles
an hour, pass each other in 30 seconds. Find the rate and
relative direction of the second train.
25. A man bought a number of sheep for £132, and having
lost 10, and sold 20 that were diseased at 6s. per head below
cost price, disposed of the remainder for XI 16, thereby
realiang his outlay. How many did he buy?
26. A boy spends 10s. in oranges and apples. The oranges
were bought at 5 for 6d., and the apples at 3 for 2d.; and
their number together amounted to 132. What did he spend
on each?
•
27. If B is allowed 2 hours more time than A takes to do
a piece of work, he will do 4 times as much, and if C is allowed
1 hour more than A, he also can do 4 times as much. More-
over, D requires 4 hours more than A to do the piece of work.
Also, the work done by A and B together is the same as that
done by C and D together in the same time. Required
the respective times for A, B, C, D to do a piece of work.
28. A person sells out £1,200 Three and a Half per Cent
stock, and invests the money in Two and a Half per Cents.,
whose price is 14 lower than the first-named stock. The loss
in annual income is £7. Find the price of the first-named
stock.
29. The banker's discount on a certain sum of money at
5 per cent, per annum is equal to the true discount on a sum
^5Q larger, Find ttie ^\xm,
SIMULTANEOUS EQUATIONS. 225
30. An express train, wliich ought to perform its journey
in 2 J hours, after having gone uniformly 80 miles, finds itself
6 minutes behind. However, by increasing the speed to as
many miles per hour as there were miles in half the journey,
it just arrived at its destination in time. Find the original
speed of the train, and the length of the journey.
31. A vessel contaiiis a quantity of spirit (sp. gr. '9) and
water, and a cylinder of wood (sp. gr. '92), whose length
is 10 inches, floats upright, so as to be just covered by the
spirit Find how much of the cylinder floats in the water.
32. A mixture of hydrogen and oxygen is found to condense
when fired, to 16 vols, of steam. Now, every 3 vols, of such
a mixture is known to condense to 2 vols., when the original
gases are in the proportion of 2 : 1. Find the quantity of
each.
33. A mixture of 1 00 grams of sodic and potassic sulphates
yielded a gram of baric sulphate. Now, each gram of sodio
sulphate yields b grams of barip sulphate, and each gram of
potassic sulphate yields c grams of baric sulphate. Find the
amount of sodic and potassic sulphates in the mixtiu:e.
34. If a oxen consume b acres of grass in c weeks, and a '
oxen consume b ' acres of grass in c ' weeks, the grass growing
uniformly, find the week's growth of an acre.
85. The freezing and boiling points of a common ther-
mometer are marked 32° and 212° respectively; those on the
centigrade thermometer are marked 0° and 100'. At whs^t
temperature do the graduations agree ?
36. A person going at the rate of a miles per hour finds
himself b hoiu^ too late when he has c miles farther to go.
How much must he increase his speed to reach home in time ?
Simultaneoos Equations of the First Degree with two
Unknown Quantities.
67. Suppose we have given the equation 3 a; - 4 y = 6,
then by ascribing to y a series of values we get a correspond-
ing series of values for x,
Thuawemayhave^ = Jj, ^ ^ 4 j, ^I,^\^>«J^^
p
226 ALGEBRA.
Again, if another equation, as 4 a? + y = 32, be given, we
may in the same way obtain a series of pairs of values which
satisfy it. And, further, if the two equations are distinct and
compatible, there is always a pair of values common to the
two eqimtions. This pair of values then satisfies both equa-
tions, and the equations are called simultaneous equations.
The methods of solving simultaneous equations will now
be explained,
PiEST Method. — Equalize the coefficients of one of the un-
Jcnown qua/atities in both equations, amd add or subtra^ct the
eqvMions so obtained, so a^ to obtain an eqv/ition with one
v/nknown qua/ntity.
Ex, 4aj + 3y = 17 (1))
9aj ^ 5y = 3 (2) / '
From (1), multiplying each side by 9, we get —
36aj + 27y = 153 (3).
And from (2), multiplying each side by 4, we get —
36aj -^ 20y = 12 (4).
(3) - (4), then 47 y = 14L
,\j/ = 3,
Hence, substituting in (1), we have—
4 a; + 3 X 3 = 17, from which we get —
aj = 2,
Hence, the solution required is a; = 2, y = 3.
Second Method. — Eocpress one of tlie uriknovyn quantities
in terms of the other by means of either equation, and sub-
stitute its value in the other.
Taking the same example, we have —
4aj + 32^ = 17
9aj - 6y = 3
From (1) we have 4 a; = 17 - 3 y, or a; = lLzJi'...(3).
Substituting this value of a; in (2) we have —
pr, 153 - 27 y - 20 y = 12, from wHch-—
y= 3.
U)-
SIMULTANEOUS EQUATIONS. 227
Then from (3) we get x = 1LZ-LL^ = 2.
Third Method. — Express cue of the wnhiown quantities
in terms of the other by means of each equ^ation, amd equate
' the expression,
"We will this time express in each case y in terms of x,
Wehave4aj + 3y = 17 (1) )
2x - 5y = 3 (2) /'
From (1) 3y = 17 - Ax, or y = 1L^^„,(3).
And from (2) 53^ = Qoj - 3, or y = ?^ " ^ (4).
Equating (3) and (4), then ^^ ~ ^'^ = ^£fl, from which
aj = 2.
Then from (3) by substitution, y = ^Lz^^JL? = 3.
Simultaneous Equations of the First Degree of Three
or more TJnlaiown Quantities.
68. In the case of three unknown quantities we may
obtain, from the three given equations, two equations
with two unknown quantities, and then, by a similar method,
from the two obtain an equation with one unknown
quantity ; and a like method may be pursued for more than
three unknown quantities.
Ex. Zx + y + 4:z =:^ 25 (1)
4a; + 3?/ - 5« = - 3 (2)
6x + 7y " 8z = 1 ^3)
(
(5)
(5)-(4),then 5y - 31« =-109 (6)
}
From(l) 12a; -h 4:y + 16« = 100 (4))
Andfrom(2) 12a; + 9 y - 15 z = - 9 (5) J
Again, from (1) 6 a; + 2y + 8z = 50 (7)
(3) - (7), then 5y - 16;s = - 49 (8)
(8) - (6), then 15 z = 60
228 ALGEBRA,
Eenoe fix)iu (8), by subslltution —
5y - 16 X 4 =-49,
from which y = 8.
And hence from (1), by substitution of the known values
of y and z —
3a;+3 + 4x4 = 25,
from which a; = 2.
Hence the respective values of a:, y, z, are 2, 3, 4.
Ex. XVIIL
1. 6 a; + y = 22, 5 a: + 3 y = 27.
2. 4 a; - 3 y = 14, G a; + 6 y = 40.
3. 3 a; + 5 y = 44, y - a; = 4.
X y__o8a y y^^ *^* '"
4. { + 3 = 2|, g + jg = 2Tfty. /^ ;:^."^
5. ? + .!^ = 4,? + ?^ = 4i. *.-: ' ^/
6^4 '4^6 * '^<<i
7. 3a; + 4y + »=ll, 2aj + y + 5» = 19,
6aj+ 2y+3«= 18.
8. 7a; + 2y+3« = 20, 3aj-4y + 2« = l,
5y-2a; + 7» = 29.
9. 2a:-5y=3, 3y-2»= -1, 4« + 2;5 = 20.
10. aaj + 6y = c, a^a? + ft^y = Cy
11. aj + y = a, y + a;=s6, aj + « = c.
12. aa; + 6y = cf, 5y + c« = e, 005 + C5 = j^.
3a;-2y 5a; + 2y ^ a: + 2y 3a;-2y_,
\. 3 + — g = 6i, — ^ 2 ^•
13.
,. aj + y aj — y_23a; + y x - y 7
^^- "IF" "*" n[5~ " 30' "IF" "*■ "To" " 30-
,^ 3 9 , 15 11
15. - + - = Ij, —
SIMULTANEOUS EQUATIONS. 229
16. 5 + >- = 5J, jr = 0.
03 — 2 y *'y a; — 2
17. 4 (a; + 3y) -3 (a; + y) = J (a; + y) (a; + 3 y),
20 7 '^
= 3.
X + 1/ fic + 3y
18. 73y - 6a5 = (05 - 6 y) (a; + 3 y),
___2 5 \; 7
X - by a; + 3y""33*
io 1 1 11.11
X y y z X z
20. 3a:-2y = 0, 4y-32j = 0, 5«-6w=14,
7u - 2a; = 3.
21. 3a;-2y = 5«-6y = 7o;-4« = l.
22. aj + y + « = a, y + » + w = 5, « + w + 05 = c,
t* + a? + y = (f .
tx^ ^ y y z ^ X z
23. - + - = a, ^ + - = 6, — + - = c.
24. 03 + ay + 6« = o, y + a« + 6aj = /3, « + oa: + 6y = y.
25. a; + ay + a'« + a' = 0.
a; + 6y + 6'« + 6' = 0.
a; + cy + €*« + c* =0,
26. X + ay + a^z + c?u + o* = 0,
a; + fty + 6'« + 6'tA + 6* = 0.
X ■\- cy '\- ^z •{■ i^u •\' c^ = 0.
a; + (fy + c?2f + cPt^ + <i* = 0.
27. a; + ay = «. 28. a ft c ,
y + (>z =: p. X y z
z + et ^ y, a' 6' c'_ j#
^ + CMC = i. X y z
u + ex ^ t, a!' h" c" .„
X y z
29. Xi + 2ajj + 3aj, + &c. + 7W?« = Oj.
a^ + 2a^ + 3aj4 + &c. +na?i =03.
&C. = &C.
«. + 2fl?i + 3aj^ + &a + naJn-x = *»•
230 ALGEBRA.
^^ C6 y ii, X y z X y z 1 1 1
aococacaoaoc
Problems producing Simultaneous Equations of the
First Degree.
69. There are certain problems which may be solved with
much greater facility by the introduction of more than one
unknown quantity.
Ex. 1. Nine men and seven women receive together
£3, 2s. 8d., and three men receive lOd. more than four women.
Find the receipts of each.
' Let X and y respectively represent in pence the receipts of
a man and woman.
Then, since £3. 2s. 8d. = 752d., expressing the conditions
of the problem in algebraical language — •
9a3 + 7y = 752) Solving these equations,
303-42/= 10 J we find —
oj = 54, y = 38.
Hence, the receipts of a man and woman are respectively
54d. and 38d., or 4s. 6d. and 3s. 2d.
Ex. 2. Eind a fraction such that^ if we diminish its
numerator by 1, it becomes equal to ^ ; and if we increase its
denominator by 1, it becomes equal to ^.
Let ^be the required fraction.
y
a; - 1 1^
Then, by problem, ^ = -- i ... .. ^ -
' "^ ^ y 7 f , which equations^ when
and _^, =l\ «^l^«<i' gi^«-
y + I bj
X = 6, y = 35.
/» .
.*. _I_ is the fraction required.
35
Ex. 3. A's money, together with twice B*s and thrice C's,
amounts to £38 ; B*s money, together with twice Cs and
thrice A'a, to £35 ; and C's money, together with twice A's
^(pfld thrice B's, also to £35. 'Em^ ^"^^ movifci ^i ^wwih.
l^ROBLEMS PRODUCING SiMtJLTANEOUS EQUATIONS. 231
Let X, y, z be respectively the number of pounds each
tas. Then we have —
y + 2» + 3a3 = 35>; from which we get —
z + 2x + 3y = 35)
05 = 5, y = 6, « = 7.
Hence, £5, £6, £7 are the respective moneys of A, B,
and C.
Ex. XIX.
1. A and B engage in play. A puts down half-a-crown
to B's florin. They play twenty games, and then it is found
that A has won 2s. How many games did each win ?
2. There is a number, the sum of whose two digits is 10,
and, if 36 be added to the number, the digits change places.
Find the number.
3. A grocer has tea at 3s. 4d. a lb., and suga|< at 4s. a stone.
He sells £2 worth of the two. If he had raised the tea 10
per cent., and lowered the sugar 12^ per cent., and sold the
same quantities of each, his profits would have been Is. 6d.
more. Find the quantity sold of each.
4. Eight times the numerator of a certain fraction exceeds
three times the denominator by 3, and five times the numer-
ator added to twice the denominator make 29. Find the
fraction.
6. If the number of cows in a field were doubled, there
would be 65 cows and horses together ; but, if the number of
horses be doubled, and that of me cows halved, there would
be 46. How many are there of each ]
6. Thirty shillings are spent in brandy, and 42s. in gin ;
19 bottles being purchased in all. Had the 42s. been spent
in brandy, and the 30s. in gin, 17 bottles only would have
been bought. Find the cost per bottle of each,
7. A fishmonger receives 240 mackerel. He sells a
certain number at 4 for a shilling, but the rest being seized
as bad fish, and he being fined 10s., finds himself a loser by
9s. Had he sold them at 3 for a shilling, he would hav^
been a gainer by 5s., if 13 more fish had been »em^. "Hiosit
many did he sell, and what did he pay for tkelo^.'^
23^ ALO£BEA«
8. Tkpee persons invest their money at 3, 4, 6 per cent-
interest respectively. The total amount of interest is £38,
and the interest of the first and third together is 2^ that of
the second ; whil^ the total interest would have been £34 had
the rates been 5, 4, 3 per cent, respectively. Find the
capital of each.
9. A toll-gate keeper receives 8s. 8d. for the toll of a
number of horses, oxen, and sheep, the tolls for each being
respectively IJd. Id., Jd. Had there been twice as many
sheep and the number of horses diminished accordingly, he
would have received 7s. 2d. Had the oxen passed through
free, and the tolls for a horse and sheep respectively been 2d.
and |d., he would have received 9s. l|d. Find the number
of each.
10. A, B, C start from the same place. B after a quarter
of an hour doubles his rate, while C, who falls, after ten
minutes diminishes his rate -^th. At the end of half an hour
A is ^ mile before B, and i mile before C, and it is observed
that liie total distance which would have been walked by the
three, had they each continued to walk imiformly from the
first, is 6^ miles. Find the original rate of each.
11. A, B, C, working 3, 4, 5 hours respectively, can do
2^f pieces of work ; if they each work an hour more, they
can finish an extra ^f of a piece; and, if does not work,
the other two, working for 1 and 6 hours respectively, can
together finish 1 piece. Find the time required for A, B,
C to finish separately a piece of work.
12. There are three numbers such that, if the first be in-
creased by 6, and the second diminished by 6, the product of
the results is the product of the first two numbers ; if the
second be increased by 2 and the third diminished by 3, the
product of the results is the product of the second and third;
and, if the first be increased by 3 and the third diminished
by 6, the product is that of the first and third. Find the
numbers.
13. A person performed a journey of 22} miles, partly by
carriage at 10 m^es an hour, partly by train at 36 miles an
hour, and the remainder by walking at 4 miles an hour. He
did the whole in 1 lioxar ^^ TKmxx\«a. "S^d he walked the
PROBLEMS PttODUCIKG SIMtfLTAKEOUS EQUATIONS. 23S
first portion, and performed the last by carriage, it -would
have taken him 2 hours 30|^ minutes. Find the respective
distances by carriage, train, and walking.
14. A and B start from two places C and D, distant 28
miles, and it is found that A reaches D 3 hours after they
meet. Had the distance between C and D been 35 miles, A
would have reached a point 28 miles from C 2 hours after
he met B. Find the respective rates of A and B.
15. Three trains — a luggage, ordinary, and express — ^move
along three parallel pairs of rails, the distance between the
stations being 120 miles. The first two start from the same
station, and the express from the opposite. The luggage
train, starting 2 hours first, is overtaken by the ordinary in
2 hours; and ^e express train, starting 1 hour after the
ordinary, meets the luggage in 1 hour 7 J minutes. Had all
three started from the same station, the ordinary would have
been overtaken in 2 hours. Find the respective rates of the
trains.
16. If (oi, biy Ci), (^2, &2> ^2)) {<hi ^s) Cj) ^G ^6 respective
compositions by weight of three mixtures of three substances,
and di, d^, d^ be llie respective prices of the mixtures, find the
price per unit of weight of each substance.
17. By alloying two ingots of gold in two given propor-
tions, we form two new ingots of which the fineness of each
is known. What is the fineness of each of the given
ingots?
18. A group of n persons play as follows: — ^The first
plays with the second and loses - of what he has, the second
then plays with the third and loses - of what he has, the
third then with the fourth, losing - of what he has, &c., the
nth with the first losing ~ of what he has. At the' end
they each have h What had each at first t
234
MATHEMATICS.
SECOND STAGE.
SECTION L
GEOMETRY.
EUCLID^S ELEMENTS, BOOK H.
Definitions.
1. A rectangle^ or right-angled parallelogram, is said to
be contained by any two of the straight lines "which contain
one of the right angles.
2. In any parallelogram, the figure which is composed of
either of the parallelograms about a diameter, together with
the two complements, is called a gnomon.
Thus the parallelogram HG, to-
gether with the complements AF,
EC, is a gnomon, which is briefly ex-
pressed by the letters AGK, or EHC,
which are at the opposite angles of
the parallelograms which make the
gnomon.
The rectangle under, or cotitaaBfe^Vs^ Vwo ^iaaiea^ as AB and BC, is
condaely expressed thus; — AB, B0»
235
FropoBition I. — TLeorem.
If there be two straight lineg, one of tohuA is divided into
any number of parte, (A« rectangle contained hy the two
straight Una is equal to the rectangles contained by the un-
divided Une, emd (Ae several parts of the divided Une.
Let A and BO be two straight lines; and let BO be
divided into any parts in the points D, E ;
The rectangle contained by the etr^ht lines A and BC A -50=
shall be equal to the rectangle contained by A and BD, a-bd +
tt^ther with that contained by A and DE, and that con- ^ " ^^ +
tained by A and EC, *'"^
CossTROCTioN. — Prom the point B draw BF at right angles
to BO (I 11),
And make EG equal to A (I. 3).
Through G draw GH parallel to
BC (I. 31),
And through the points D,E,0,di'a'w .
DK, EL, CH pawUel to BG (I. 31).
Proof. — Then the rectangle BH is ,
equal to the rectangles £K, DL, EH.
But BH is contained by A and BC, for it is contained ^■
by GB and BC, and GB is equal to A (Const) ;
And BK is contained by A end BI>, for it is cont^ned
by OB cmd BD, and GB is equal to A;
And DL is contained by A and DE, because DK is equal
to BG, which is equal to A (L 34) ;
And in like manner EH is conttdned by A and KC ;
Therefore the rectangle contained by A and BO is equal
to the several rectangles contained by A and BD, by A and
DE, and by A and EC.
Therefore, if there be two straight lines, &£. Q. E. D.
PropoBition S. — Theorem.
If a etraight line be divided into any two parts, the rect-
angles contained hy the whole line and each of its parts are
together egual to the equare on the whole line.
Let tlie straight line AB be divided into any tvo ^nsVs^
in Uie point C ;
I JL
236
GEOMETRY.
AB'BC
-4-AB-AO
= AB?.
Por
ABS is the
sum of its
parts AF+
CE.
A C B
n J^ B
And;.=
AB-AC+
AB'BC.
The rectangle contained by AB and BC, together with it a
rectangle contained by AB and AC, shall
be equal to the square on AB.
Construction. — Upon AB describe the
square ADEB (I. 46).
Through C draw CF parallel to AD or
BE (I. 31).
Proof. — Then AE is equal to the rect-
angles AF and CE.
But AE is the square on AB ;
Therefore the square on AB is equal to the rectangles AF
and CE.
And AF is the rectangle contained by BA and AC, for it
is contained by DA and AC, of which DA is equal to BA;
And CE is contained by AB and BC, for BE is equal
to AB.
Therefore the rectangle AB, AC, together with the rect-
angle AB, BC, is equal to the square on AB.
Therefore, if a straight line, &c. Q, E, D,
AB-BC=
BC«+
ACBC.
For
AE=:AD
+CE.
Proposition 3. — Theorem.
If a etraigld line he divided into amy tu>o parts, the rect-
angle contained hy the wlioh and one of the pa/rt8 is equal to
the square on that pa/rty together vdth the rectam^gle contained
ky the two parts.
Let the straight line AB be divided into any two parts in
the point C;
The rectangle AB * BC shall be eq\ial to the square on BC,
together with the rectangle AC • CB.
Construction. — Upon BC describe the square CDEB
(I. 46).
Produce ED to F; and through A
draw AF paraUel to CD or BE (I. 31).
Proof. — ^Then the rectangle AE is
equal to the rectangles AD and CE.
But AE is the rectangle contained by
AB and BC, for it is contained by AB
and BE, of which BE is equal to BC ;
And AD is contameA\jy AC ^cid CB, for CD is equal to CB;
PROPOSITIONS. 237
And CE is the square on BC.
Therefore the rectangle AB, BC is equal to the square on
BC, together with the rectangle AC, CB.
Therefore, if a straight line, &c. Q. E, D.
Proposition 4.— Theorem.
If a straight line he divided into any two parts^ the square
on the wJioJs line is equal to tJie sum of tlie squares on the two
parts, together with tunce the rectangle contained by the parts.
Let the straight line AB be divided into any two parts
inC;
The square on AB shall be equal to the squares on AC aaid ab« =
CB, together with twice the rectangle contained by AC ^tJ^l
and CB. "^'^^^
CoNSTBUCTiON. — ^Upon AB describe the square ADEB
(I. 46), and join BD.
Through C draw CGF parallel to AD or
BE (I. 31).
Through G draw HGK parallel to AB " ■
or DE (I. 31).
Proof. — Because CF is parallel to AD,
and BD falls upon them,
Therefore the exterior angle BGC is
equal to the interior and opposite angle ADB (I. 29).
Because AB is equal to AD, being sides of a square, the
angle ADB is equal to the angle ABD (I. 5);
Therefore the angle CGB is equal to the angle CBG Show first
^-o-a- xy, a square
Therefore the side BC is equal to the side CG (I. 6). = cb^-
But CB is also equal to GK, and CG to BK (I. 34);
Therefore the figure CGKB is equilateral.
It is likewise rectangular.
For since CG is parallel to BK, and CB meets them, the
angles KBC and GCB are together equal to two right
angles (I. 29).
But ElBC is a right angle (Const.), therefore GCB is a
right angle (Ax. 3).
Therefore also the angles CGK, GKB, oi^t^svX^ \^ ^^'efc^
are nght angles (L 34),
A
C Tt
I °
/
p
1
K
238 GEOMETBT.
Therefore CGKB is rectangular; and it has been proved
equilateral; therefore it is a square; and it is upon the
sideCB.
Co aiao ^^^ ^^6 same reason HF is also a square, and it is on
11F=AC8 the side HG, which is equal to AC (I. 34).
Therefore HF and CK are the squares on AC and CB.
And because the complement AG is equal to the com-
^„j plement GE (I. 43),
AO+GE And that AG is the rectangle confecined by AC and CO,
=2AC-CB. ^at is, by AC and CB,
Therefore GE is also equal to the rectangle AC, CB ;
Therefore AG, GE are together equal to twice the rect-
angle AC, CB;
And HF, CK are the squares on AC and CB.
Therefore the four figures HF, CK, AG, GE are equal
to the squares on AC and CB, together with twice the
rectangle AC, CB.
But HF, CK, AG, GE, make up the whole figure ADEB,
which is the squai'e on AB ;
.'.whole Therefore the square on AB is equal to the squares on
^^re^or ^Q ^^^ Q^ ^^^ twice the rectangle AC • CB.
+2 Ac^: Therefore, if a straight line, &c. Q, E. D.
Corollary. — From this demonstration it follows that the
parallelograms about the diameter of a square are likewise
squares.
Proposition 5. — Theorem.
If a straight line he divided into two eqtud parts, and also
into two uneqiuil pa/rts, the rectangle contained by the uneqtial
parts, together with the sqiux/re on the line between tfie points of
section, is equal to the square on IwJf the line.
§
Let the straight line AB be bisected in C, and divided
unequally in D ;
AD-DB The rectangle AD, BD, together with the square on CD,
icgj shall be equal to the square on CB.
Construction. — ^Upon. CB describe the square CEFB
(X 46), and join BE,
PBOPOSITIONS.
239
Through D draw DUG parallel to CE or BF (I. 31).
Through H draw KLM parallel to CB or EF.
And through A draw AK par-
allel to CL or BM. a. c d b
Proof. — Then the complement
CH is equal to the complement HF fe"
(I. 43).
To each of these add DM ; there-
fore the whole CM is equal to the
whole DF (Ax. 2).
But CM is equal to AL (I. 36), because AC is equal to CB
(Hyp.);
Therefore also AL is equal to DF (Ax. 1).
To each of these add CH; therefore the whole AH is
equal to DF and CH (Ax. 2).
But AH is contained by AD and BD, since DH is equal
to DB (II. 4, cor.),
And DF, together with CH, is the gnomon CMG ;
Therefore the gnomon CMG is equal to the rectangleAD, DB.
To each of these equals add LG, which is equal to the
square on CD (II. 4, cor., and I • 34);
Therefore the gnomon CMG, together with LG, is equal
to the rectangle AD, DB, together with the square on CD.
But the gnomon CMG and LG make up the whole figure
CEFB, which is the square on CB;
Therefore the rectangle AD, DB, together with the square
on CD, is equal to the square on CB.
Therefore, if a straight line, &c. Q, U. D,
GoBOLLART. — From this proposition it is manifest that the
difference of the squares on two unequal lines AC, CD is
equal to the jrectangle contained by their sum and difference.
For
AL=CM
=DF.
.-. AH=
DF+CH.
.•.CMG.
=ADDB
Add to
each LG o
CD*.
.'. CB«
=ADDB
+CD«.
Proposition 6.— Theorem.
If a straight line he bisectedf cmd produced to any pointy the
reda/ngle contained hy the whole line thus produced and the
paH cf U produced, together with tlie squa/re on half the line
hiaeded^ is eqvxd to tJie square on the straight line which is
Tnade up of the half ami the part produced.
Let the Btr&ightlmeAB be bisected inC^andpToAuce^XjoT^ \
210
AD-im
+ CBS
Tor
= HF.
.'.AM or
AD-DB
= CMa
Addto
eacbLO
orCB*.
.-.AD'DB
-i-OBS
A
C B B
L
H
/
K
/
Tbe rectangle AD, DB, together with the square on CB,
shall be equal to the sqnare on CD.
Construction. — ^TJpon CD describe tbe sqoare C£FD
(L 46), and join DK
ThnHi^ B draw BHG parallel
to CE or DF (L 31).
Throng H draw KLM parallel
to AD or EF.
And throogh A draw AK par-
allel to CL or DM.
Proof. — Because AC is eqnal
to CB (Hyp.), the rectangle AL is eqnal to CH (L 36).
But CH is eqnal to HF (L 43), therefore AL is equal to
HF (Ax. 14).
To each of these add CM; therefore the whole AM is
equal to the gnomon CMG (Ax. 2).
But AM is the rectangle contained by AD and DB, since
DM is equal to DB (n. 4, cor.) ;
Therefore the gnomon CMG is equal to the rectangle
AD, DB (Ax. 1).
Add to each of these IXx, which is equal to the square on
CB (n. 4, cor., and L 34) ;
Therefore the rectangle AD, DB, together with the square
on CB, is equal to the gnomon CMG and the figure LG.
But the gnomon CMG and LG make up the whole figure
CEFD, which is the square on CD;
Therefore the rectangle AD, DB, together with the square
on CB, is equal to the square on CD.
Therefore, if a straight line, <&c. Q. K D.
Proposition 7. — Theorem.
If a straight line he divided into amy two parts, the squares
on the wliole line a/nd on one of the pa/rts are equal to twice
the rectangle contained hy tlie whole and that pa/rt, together
with the square on the otlier part.
Let the straight line AB be divided into any two parts in
the point C;
^^^^ 2^0 squares on AB and BC shall be equal to twice the
y- AC*, frect&ngle AB, BC, toge\iv^x m^ \5afe ^opsca oa. AC!^
PROPOSITIONS.
241
For
AK=CE.
CoNSTRUOTiOK. — Upon AB describe the square ADEB
(I. 46), and join BD.
Through C draw CGF parallel to AD or
BE (I. 31).
Through G draw HGK parallel to AB
orDE(L 31).*
Proof. — ^Then AG is equal to GE
(143).
To each of these add CK ; therefore the
whole AK is equal to the whole CE;
Therefore AK and CE are double of AK.
But AK and CE are the gnomon AKF, together with the
square CK ;
Therefore the gnomon AKF, together with the square CK,
is double of AK.
But twice the rectangle AB, BC is also double of AK, for
BK is equal to BC (II. 4, cor.);
Therefore the gnomon AKF, together with the square CK,
is equal to twice the rectangle AB, BC.
To each of these equals add HF, which is equal to the Add hf or
square on AC (II. 4, cor., and I. 34); t^T"^^
Therefore the gnomon AKF, together with the squares ^ *
CK and HF, is equal to twice the rectangle AB,* BC, together
with the square on AC.
But the gnomon AKF, together with the squares CK and .*• ab8
HF, make up the whole figure ADEB and CK, which are ieABBC
the squares on AB and BC; +ac2.
Therefore the squares on AB and BC are equal to twice
the rectangle AB, BC, together with the square on AC.
Therefore, if a straight line, &c. Q. K D,
.-.AKF
+ CK =
2ABBC.
Proposition 8. — Theorem.
If a straight line he divided into any two pa/rt8ff<mr times
the recta/ngle contained by tJie whole line and one o/* the parts ,
together with the square on the other part, is equal to the
gqtuire on the straight line which is made up of the whole line
cmd thejvrst mentioned part.
Let the straight line AB be divided into' any two partsi iix
the point C;
Si3 GBOVETBT.
<AB^BC Fonr times the rectangle AB, BC, bigeUier with Uie square
<A£t-BQ:. oa AX), shall be equal to the square on the straight Ime
made up of A£ and BC together.
CosOTsrerios. — Produce AB to D, so that SD may be
equal to CB (Poet 3. and L 3).
Upon AD describe the square AEFD
(I- 46),
And constnict two figures such as in " —
the preceding propositjoas, ^
Pboof.— Becanse CB ia equal to BD
(Const), CB to GK, and BD to KN
For the aagae reason PR is equal
toRO.
r„ And because CB is equal to BD, and GK to KN,
SN=CK Iherefore the Eectangle CK is equal to BN, and GB to
But CK is equal to BN, because they are the complements
of the parallelogram CO (L 43) ;
Therefore also BN is equal to GR (Ax. 1).
.-, iheloor Therefore the four ractongles BN, CK, GR, RN are equal
^CK to one another, and so the four are quadruple of one of them,
■ CK
^^ Again, because CB is equal to BD (Const);
t^ "ct- And that BD is equal to BK, that is CG (IL 4, Cor., and
^%., 1.34);
And that CB is equal to GK, that is GP (L 34, and IL 4,
„^
/
r
A
k
—
r^""
Therefore CG is equal to GP (Ax. 1).
And because CG is equal to GP, and PR to RO,
The rectangle AG is eqnal to MP, and PL to RF (L 36).
But MP is equal to PL, because they are complements of
the parallelogram ML (L 43), and AG is equal to RF (Ax. 1);
Therefore the four rectangles AG, MP, PL, RF are equal
to one another, and so the four are quadruple of one of them,
'AG.
And it was demonstrated that the four CK, BN, GB, and
BN are quadruple of CK j
3^erefore the eight rectuigles which make np the gnomon
AOS. are quadruple cS AS.
PROPOSITIONS. 243
And becauscf AK is the rectangle contained by AB and .-.Gnomo
BC, for BK is equal to BC ; ^ck^c
Therefore four times the rectangle AB, BO is quadruple of ^1^.«
But the gnomon AOH was demonstrated to be quadruple
of AKj
Therefore four times the rectangle AB, BC is equal to the
gnomon AOH (Ax. 1).
To each of these add XH, which is equal to the square Hence
on AC (11. 4, cor., and I. 34); ^^^^^o
Therefore four times the rectangle AB, BC, together with
the square on AC, is equal to the gnomon AOH and the
square XH.
But the gnomon AOH and the square XH make up the
figure AEFD, which is the square on AD ;
Therefore four times tljie rectangle AB, BC, together with * abbc
the square on AC, is equal to the square on AD) that is, on = af =
the line made up of AB and BC together. (ab+bc)2
Therefore, if a straight line, &c. Q, E. D.
Proposition 9.— Theorem.
If a straight line he divided into two equal, and also into
two unequal parts, the sgiuires on tJie two unequal pa/rts are
together double of the Square on JiaJfthe line, and of the square
on tlie line between the points of section.
Let the straight line AB be divided into two equal' parts
in the point C, and into two unequal parts in the point D ;
The squares on AD and DB shall be together double of ads+dbs
the squares on AC and CD. cM^^^^
CoirsTRUCTiON.— From the point C draw CE at right
angles to AB (1. 11), and make it equal
to AC or CB (I. 3), and join EA, EB.
Through D draw DF parallel to CE
(I. 31).
Through F draw FG parallel to BA
(L 31), andjomAF.
Peoop. — ^Because AC is equal to CE
(Confit), ihe angle EAC ia equal to the angle AEC ij.. ^"^^
«
244 GEOXETET.
And because the angle ACE is a right angie (Const.), the
angles A EC and EAC together make one right angle (L 32),
For and they are equal to one another ;
A^=i * Therefore each of the angles A EC and EAC is half a right
= CEB. angle.
For the same reason, each of the angles CEB and EBC is
half a right angle;
'^i^? ^ Therefore the whole angle AEB is a right angle.
And because the angle GEE is half a right angle, and the
angle EGF a right angle, for it is equal to the interior and
opposite angle ECB (L 29),
Therefore the remaining angle EFG is half a right angle ;
Therefore the angle GEF is equal to the angle EFG, and
EG=OF. the side EG is equal to the side GF (L 6).
Again, because the angle at B is half a right angle, and
the angle FDB a right angle, for it is equal to the interior
and opposite angle ECB (I. 29),
Therefore the remaining angle BFD is half a right angle ;
^^ Therefore the angle at B is equal to the angle BFD, and
DF=DB. the side DF is equal to the side DB (I. 6).
And because AC is equal to CE (Const.), the square on
AC is equal to the square on CE ;
Therefore the squares on AC and CE are double of the
But square on AC.
A^= But the square on AE is equal to the squares on AC and
CE, because the angle ACE is a right angle (I. 47);
Therefore the square on AE is double of the square on AC.
Again, because EG is equal to GF (Const), the square on
EG is equal to the square on GF ;
Therefore the squares on EG and GF are double of the
square on GF.
Soaiso But the square on EF is equal to the squares on EG and
=2 cb«. GF, because the angle EGF is a right angle (I. 47) ;
Therefore the square on EF is double of the square on GF.
And GF is equal to CD (I. 34);
Therefore the square on EF is double of the square on CD.
But it has been demonstrated that the square on AE is
• Ajiyj. ^^ double of the square on AC ;
iFv= Therefore the squares ox\ AB! W^d BF are double of the
^^ equeffeu o^ AO m^ CD,
PROPOSITIONS.
245
But the sqiiare on AF is equal to the squares on AE and
EF, because the angle AEF is a right angle (I. 47) ;
Therefore the square on AF is double of the squares on AC
and CD.
But the squares on AD and DF are equal to the square
on AF, because the angle ADF is a right angle (I. 47);
Therefore the squares on AD and DF are double of the
squares on AC and CD.
And DF is equal to DB; therefore the squares on AD and
DB are double of the squares on AC and CD.
Therefore, if a straight line, &c. Q,E,D,
But
AE2+EF«
=AFa
=ADa+
DF«
=ADa+
DB*.
.•.AI>«+
DB
=2fAC24
CD2)
Proposition 10. — Theorem.
If a straight line he bisected and produced to any pointy the
sqiuj/re on the whole line thus produced^ and tJie square on the
part of it produced y a/re together double of the square on lialf
the line bisected, ami of the sqtcare on tlie line made up of the
half and the pa/rt produced.
Let the straight line AB be bisected in C, and produced to D ; f^^+^^^^
The squares on AD and DB shall be together double of the cb*).
squares on AC and CD.
Construction. — ^From the point C draw CE at right angles
to AB, and make it equal to AC or CB (I. 11, 1. 3), and join
AE, EB.
Through E draw EF parallel to AB, and through D draw
DF parallel to CE (I. 31). Then because the straight line
EF meets the paraUels EC, FD, the angles CEF, EFD are
equal to two right angles (I. 29); therefore the angles BEF,
EFD are less than two right angles;
therefore EB, FD will meet, if
produced towards B and D (Ax.
12).
Let them meet in G, and join a^
AG.
Proof. — Because AC is equal
to CE (Const.), the angle CEA is equal to the angle EAC
(L6).
And the angle ACE is a right angle ; therefor© eaftbL^l^Jckft
angle» CEA and JEAO ia half a right angle (1. ^^Y
24G OEOMETBT.
As in For the same reafion each of the angles CEB and EBC is
^^^' ' half a right angle ;
ri^ht*z* Therefore the whole angle AEB is a right angle.
And because the angle EBC is half a right angle, the angle
DBG, which is vertically opposite, is also half a right angle
But the angle BDG is a right angle, because it is equal to
the alternate angle DCE (I. 29);
. Therefore the remaining angle DGB is half a right angle,
and is therefore equal to the angle DBG;
An<]_jjQ Therefore also the side BD is equal to the side DG (I. G).
Again, because the angle EGF is half a right angle, and
the angle at F a right angle, for it is equal to the opposite
angleECD(L 34);
Therefore the remaining angle FEG is half a right angle
(I. 32), and therefore equal to the angle EGF;
Aiso^ Therefore also the side GF is equal to the side FE (I. 6).
" ' And because EC is equal to CA, the square on EC is equal
to the square on CA;
Therefore the squares on EC and CA are double of the
square on CA.
Again as But the squaipo on AE is equal to the squares on EC and
in Prop. 9. (.^ ^j^^y^
AE2= Therefore the square on AE is double of the square
^^^'' on AC.
Again, because GF is equal to FE, the square on GF is
equal to the square on FE;
Therefore the squares on GF and FE are double of the
square on FE.
But the square on EG is equal to the squares on GF and
EE (I. 47) ;
Therefore the square on EG is double of the square on FR
And FE is equal to CD (I. 34),
Ami Tlicreforo the square on EG is double of the square on CD.
2^CD3r ^^^* i* ^^ ^®®^ demonstrated that the square on AE is
.•.AE"^+ double of the square on AC;
2(AC2+ Therefore the squares on AE and EG arc double of the
CJi^^)' squares on AC and CD.
aM+eg3 -^"^ *^^ square on AG is c(\ual to the squares on AE and
^AQJ. EG (I. 47);
PROPOSITIONS.
247
Therefore the square on AG is double of the squares on /.ag*
AC and CD. d^'"^
But the squares on AD and DG are equal to the square on .•.ad^+
AG (I. 47); ^^AC+
Therefore the squares on AD and DG are double of the cd«).
squares on AC and CD.
And DG is equal to DB; therefore the squares
on AD and DB are double of the squares on AC
and CD.
Therefore, if a straight line, &c. Q,E.D.
Proposition 11.— Problem.
To divide a given straight line into two parts, so that the
rectangle contained by the whole and one of tJie parts sliall he
equal to the square on tlie otfier part
Let AB be the given straight line.
It is required to divide AB into two parts, so that the
rectangle contained by the whole and one of the parts shall
be equal to the square on the other part.
Construction. — Upon AB describe
the square ABDC (I. 46).
Bisect AC in E (1. 10), and join BE.
Produce CA to F, and make EF equal
to EB (I. 3).
Upon AF describe the square AFGH
(I. 46).
Produce GH to K.
Then AB sliaU he divided in H, so
tJuU the rectamgle AB, BHis eqv<dto tlie
square on AH.
Proof. — Because the straight line AC is bisected in E,
and produced to F,
The rectangle CF,TFA, together with the square on AE, cffa +
is equal to the square on EF (II. 6). 4ef«
But EF is equal to EB (Const.); =eb»
Therefore the rectangle CF, FA, together with the square
on AE, is equal to the square on EB.
But the square on EB is equal to the squares o\x KEi ^sA ^Ks^v
AB, because tho smgh EAB is a right anglo (J. Vl"^', ^*^^*
248
OEOMETRl?.
CFFA
=AB«.
Therefore the rectangle CF, FA, together with the square
on A£, is equal to the squares on AE and AB.
Take away the square on AE, which is common to both ;
Therefore the remaining rectangle CF, FA is equal to the
square on AB (Ax. 3).
But the figure FK is the rectangle contained by CF and
FA, for FA IS equal to FG;
And AD is the square on AB;
.FK=AD Therefore the figure FK is equal to AD.
Take away the common part AK, and the remainder FH
is equal to liie remainder HD (Ax. 3).
But HD is the rectangle contained by AB and BH, for
A.B is equal to BD;
And FH is the square on AH ;
Therefore the rectangle AB,BH is equal to the square on
AH.
Therefore the straight line AB is divided in H, so that the
rectangle AB, BH is equal to the square on AH. Q.E.F,
Take away
AK. then
FH=HD
or AB-BH
=AH«.
For
AB2=ACa
+CB«+
2BCCD.
BD2=BC2
+CD«+
2BC-CD.
Proposition 12.— Theorem.
In obtuse-angled triangles if a perpendicular he drawn from
eitlier of the acute angles to the opposite side prodticed, tite
square on the side subtendiTfg the obtuse angle is greater than
tJie squa/res on the sides containing the obtuse angle, by twice
the recta/ngle contained by tlie side on which, wlien produced^
the perpendicula/r faUs, amd the straight lins intercepted with-
out the triangle, between ihe perpendicular and t/ie obtuse angU,
Let ABC be an obtuse- angled triangle, having the obtuse
angle ACB; and from the point A let AD
be drawn perpendicular to BC produced.
The square on AB shall be greater
than the squares on AC and CB by twice
the rectangle BC, CD.
Proof. — ^Because the straight line BD
is divided into two parts in tiie point C,
The square on BD is equal to the
squares on BC and CD, and twice the rectangle BO, CD
{II. 4).
To each of these ecfoaXa aM ^a wsfvasft Qti DA ;
PROPOSITIONS. 249
Therefore the squares on BD and DA are equal to the .'.bdhd.
squares on BC, CD, DA, and twice the rectangle BC, CD. =bc«+
But the square on BA is equal to the squares on BD and ffK"^
DA, because the angle at D is a right angle (I. 47) ; +2BCCE
And the square on CA is equal to the squares on CD and X§^'^
DA (I. 47) ; +2BCcr
Therefore the squai'e on BA is equal to the squares on BC
and CA, and twice the rectangle BC, CD ; that is, the square
on BA is greater than the squares on BC and CA by twice
the rectangle BC, CD.
Therefore, in obtuse-angled triangles, &c. Q.KD.
Proposition 13.— Theorem.
In every triangle, tlie sqiuire on tlie side subtending an acute
angle is less tha/n the sqiuires on tlve sides containing that angle,
hy ttcice the rectangle contain^ by either of these sides, and
the straight line intercepted between the perpendicular let fall
on it from, tlie opposite angle and tlie acute angle.
Let ABC be any triangle, and the angle at B an acute
angle ; and on BC, one of the sides containing it, let fall the
perpendicular AD from the opposite angle (I. 12).
The square on AC, opposite to the angle B, shall be less ac2=cb'
than the squares on CB and BA, by twice the rectangle ^b?bd
CB, BD.
Case L— First, let AD fall within the
triangle ABC.
Proof. — Because the straight line CB is
divided into two parts in the point D,
The squares on CB and BD are equal / \ cbVbd
to twice the rectangle contained by CB, ^ ^ -^ +dc?^^
BD, and the square on DC (II. 7).
To each of these equals add the square on DA.
Therefore the squares on CB, BD, DA are equal to twice .•.CB«+
the rectangle *CB, BD, and the squares on AD and DC. S5)^
But the square on AB is equal to the squai-es on BD and Tff ^ff
DA, because the angle BDA is a right angle (I. 47) ; D02),
And the square on AC is equal to the squares on AD and cb2+ba'
DC (L 47) ; =2cb;d;
Therefore the squares on CB and BA. ate ec^a^. \o "^"a
250
GEOMETRY.
.•.AC*-«
=CB2
+ BA2 -..
2CBBD.
AB2=ACa
+CB2+
BCCD.
.•.AB2+
BC3
=AC2+
2(BCS4-
BCCD).
Now
DB-BC=
BCCD+
BC3.
.-.ACa
=AB2+
BC2-
2BCDB.
square on AC, and twice the rectangle CB, BB; that is, the
square on AC alone is less than the squares on CB and BA
by twice the rectangle CB, BD.
Case II. — Secondly, let AJD fell without the triangle ABC.
Proof. — Because the angle at D is a right angle (Const.),
^ the angle ACB is greater than a right
angle (I. 16);
Therefore the square on AB is equal
to the squares on AC and CB, and twice
the rectangle BC, CD (II. 12).
To each of these equals add the square
^ on BC.
Therefore the squares on AB and BC are equal to the
square on AC, and twice the square on BC, and twice the
rectangle BC, CD (Ax. 2).
But because BD is divided into two parts at C,
The rectangle DB, BC is equal to the rectangle BC, CD
and the square on BC (II. 3) ;
And the doubles of these are equal, that is, twice the
rectangle DB, BC is equal to twice the rectangle BC, CD
and twice the square on BC ;
Therefore the squares on AB and BC are equal to the
square on AC, and twice the rectangle DB, BC ; that is, the
X square on AC alone is less than the squares on AB
and BC by twice the rectangle DB, BC.
Case III. — ^Lastly, let the side AC be perpen-
dicular to BC.
Proof. — Then BC is the straight line between the
perpendiciUar and the acute angle at B ; and it is
"c manifest that the squares on AB and BC are equal
to the square on AC, and twice the square on BC (I. 47,
and Ax. 2).
Therefore, in every triangle, &c. Q.KD.
Proposition 14.— Problem.
To describe a square tJiat elialt he equal to a given rectu
lineal figure*
Let A bo the given rectilineal figure.
It 18 required to deBCTib© a^c^vvax^ \i5a».\» ^Joalll be equal to A.
PROPOSITIONS.
251
Construction. — Describe the rectangular parallelogram
BCDE equal to the rectilineal figure A (I. 45).
If then the sides of it, BE, ED, are equal to one another
it is a square, and what
was required is now
done.
But if they are not
equal, produce one of
them, BE, to F, and
make EF equal to ED
(I. 3).
Bisect BF in G (L 10), and from the centre G, at the dis-
tance GB, or GF, describe the semicircle BHF ;
Produce DE to H, and join GH ;
TJien the square described upon EH sltall he equal to tlie
rectilineal figure A,
Proof. — ^Because the straight line BF is divided into two
equal parts in the point G, and into two unequal parts in
the point E,
The rectangle BE,EF, together with the square on GE, be-ef
is equal to the square on GF (II. 5). ^^pj
But GF is equal to GH : =oh«
Therefore the rectangle BE,EF, together with the square ehs.
on GE, is equal to the square on GH.
But the square on GH is equal to the squares on GE and
EH (I. 47) ;
Therefore the rectangle BE,EF, together with the square .-.be-ef
on GE, is equal to the squares on GE and EH. =^2.
Take away the square on GE, which is common to both ;
Therefore the rectangle BE,EF is equal to the square on
EH (Ax. 3).
But the rectangle contained by BE and EF is the paral-
lelogram BD, because EF is equal to ED (Const.);
Therefore BD is equal to the square on EH.
But BD is equal to the rectilineal figure A (Const) ;
Therefore the square on EH is equal to the rectilineal Hence
figure A. EH2=A.
Therefore, a square has been made equal to the given
rectilineal figure A, viz., the squai-e described on EH,
Q,E.F.
252 GEOMETttV.
EXERCISES ON BOOK II.
Prop. 1—11.
1. Divide a given straight line into two sncli parts that the rect*
angle contained by them may be the greatest possible.
2. The sum of the squares of two straight lines is never less than
twice the rectangle contained by the straight lines.
3. Divide a given straight line into two parts such that the squares
of the whole Ime and one of the parts shall be equal to twice the
square of the other part.
4. Given the sum of two straight lines and the difference of their
squares, to find the lines.
5. In any triangle the difference of the squares of the sides is equal
to the rectangle contained by the sum ana difference of the parts
into which the base ia divid^ by a perpendicular from the vertical
angle.
6. Divide a given straight line into such parts that the sum of their
squares may be equal to a given square.
7. If ABOD be any rectangle, A and C being opposite angles, and
O any point either within or without the rectangle — OA* + OC*
= 0B« + 0D«.
8. Let the straight line AB be divided into any two parts in the
point C. Bisect OB in D, and take a point E in AC such that EC
= CD. Then shall AD> = AE« + AC • CB.
9. If a point C be taken in AB, and AB be produced to D so that
BD and AC are equal, show that the squares described upon AD and
AC together exceed the square upon AB by twice the rectangle con-
tained by AE and AC.
10. From the hypothenuse of a right-angled triangle portions are
cut off equal to the adjacent sides. Show that the square on the
middle segment is equal to twice the rectangle under the extreme
segments.
11. If a straight line be divided into any number of parts, the
square of the whole line is equal to the sum of the squares of the
parts, together with twice the rectangles of the parts taken two and
two together.
12. If ABC be an isosceles triangle, and DE be drawn parallel to
the base BC, cutting in D and E either the side or sides produced,
and EB be joined; prove that BE« = BC • DE + CE*.
Prop. 12—14.
13. In any triangle show that the sum of the squares on the sides
is equal to twice the square on half the base, and twice the square an
the line drawn from tne vex\iex to t\i^ middle of the base.
EXERCISES ON THE PROPOSITIONS. 253
14. If squares are described on the sides of any triangle, find the
difference oetween the sum of two of the s(][uares and the third
square, and show from your result what this becomes when the
angle opposite the third square is a right angle.
15. Show also what the difference becomes when the vertex of the
triangle is depressed until it coincide with the base.
16. The square on any straight line drawn from the vertex of an
isosceles triangle, together with the rectangle contained by the
segments of the base, is equal to the square upon a side of the
triangle.
17. If a side of a triangle be bisected, and a perpendicular drawn
from the middle point of the base to meet the side, then the square
of the altitude of the triangle exceeds the square upon half the base
by twice the rectangle contained by the side and the straight line
between the points of section of the side.
18. In any triangle ABC, if perpendiculars be drawn from each of
the angles upon the opposite sides, or opposite sides produced, meet*
ing them respectively m D, E, F, show that —
BA« + AC« + CB« = 2AE • AC + 2CD -CB + 2BF BA;
all lines being measured in the same direction round the triangle.
19. Construct a square equal to the sum of the areas of two given
rectilinear figures.
20. The base of a triangle is 63 ft., and the sides 25 ft. and 52 ft.
respectively. Show that the segments of the base, made by a per-
pendicular from the vertex, are 15 ft. and 48 ft. respectively, and
that the area of the triangle is 630 sq. ft.
21. In the same triangle, show that the length of the line joining
the vertex with the middle of the base is 22*9 ft.
22. A ladder, 45 ft. long, reaches to a certain height against a
wall, but, when turned over without moving the foot, must be short'
ened 6 ft. in order to reach the same height on the opposite side.
Supposing the width of the street to be 42 ft., show that the height
to -vmich the ladder reaches is 36 ft.
23. The base and altitude of a triangle are 8 in. and 9 in. respec-
tively; show that its area is ecyasl to a square whose side is 6 in.
Prove your result by construction.
24. On the supposition that lines can be always expressed exactly
in terms of some unit of length, what geometrical propositions may
be deduced from the following algebraical identities ? —
(1.) (a + 6)« = a« + 2ab + 6«
(2.) {a + b){a-b) + b^ = a«
(3.) (2a + 6)6 + a« = (a + 6)»
(4.) (a + 6)« + 6« = 2 (a + 6) 6 + a^
(5.) 4(a + b)b + b^ = {a + 2b)^
(6.) (a + 6)« + (a - 6)« = 2a« + 2ir'
(7.) (2a + 6)« + 6> = 2aa + 2 (<» + ir
254
PEoionrRT,
EUCUD'S ELEMENTS, BOOK UL
Peflnitions.
1 . Equal circles are tliose of which the diameters are equal,
or from the centres of which the straight lines to the circum<
ferences are equal.
2. A straijpht line is said to toucli
a circle, or to be a tangent to it,
when it meets the circle, and being
produced does not cut it.
3. Circles are said to touch one
another, which meet, but do not cut
one another.
4. Straight lines are said to be equally
distant from the centre of a circle, when
the perpendiculars drawn to them &om the
centre are equal.
5. And the straight line on which the
greater perpendicular falls, is said to be
farther from the centre.
6. A segment of a circle is the figore
contained by a straight line and the circum-
ference which it cuts off.
7. The angle of a segment is that which is contained by
the straight line and the circumference.
^^_^^^ 8. An angle in a segjment is the angle
/\. ^\^ contained by two straight lines drawn
/ \ \ from any point in the circumference of
I \ I the segment to the extremities of the
\ \ / straight line, which is the base of the
segment.
9. An angle is said to insist or
stand upon the circumference inter-
cepted between the straight lines that contain the
tangle.
PROPOSITIONS.
255
10. A sector of a circle is the figure con-
tained by two straight lines drawn from
the centre and the circumference between
them.
11. Similar segpoients of circles
are those which contain equal
angles.
[Any portion of the circiunference is called an arc, and the chord
of w arc is the straight line joining its extremities.]
Proposition 1. — Problem.
> To finJ, the centre of a given circle.
Let ABC be the given circle.
It is required to find its centre.
Construction. — Draw within the circle any chord AB,
and bisect it in D (I. 10).
From the point D draw DC at right
angles to AB (I. 11).
Produce CD to meet the circumference
in E, and bisect CE in F (I. 10).
Then the point F slwll he tlie centre of the
circle ABC,
Proof. — For if F be not the centre, if ^'
possible let G be the centre j and join GA,
GD, GB.
Then, because DA is equal to DB (Const.), and Dvr com-
mon to the two triangles ADG, BDG;
The two sides AD, DG are equal to the two sides BD,
DG, each to each ;
And the base GA is equal to the base GB, being radii of
the same circle;
Therefore the angle ADG is equal to the angle BDG (I. 8). .^. z adg
But when a straight line, standing on another straight ~
line, makes the adjacent angles equal to one axi.o\^\i<^T, ^%j^ <^1
the toiglea is called a right angle (I. Def. lOy
Suppose
Othe
centre.
256 GEOMETRY.
Therefore the angle GDB is a right angle.
But the angle FDB is also a right angle (Const.) ;
£:^Ggg Therefore the angle GDB is equal to the angle FDB
"" * (Ax. 11), the less to the greater; which is impossible.
Therefore G is not the centre of the circle ABC.
In the same manner it may be shown that no point which
is not in CE can be the centre.
And since the centre is in CE, it must be in F, its point
of bisection.
Therefore F is the centre of the circle ABC : which was
to be found.
Corollary. — From this it is manifest that, if in a circle a
straight line bisect another at right angles, the centre of the
circle is in the line which bisects the other.
Proposition 2. — Theorem.
If any two points be taken in the circumference of a circle^
iJie straight line which joins them shoM/aU vnthin the circle.
Let ABC be a circle, and A and B any two points in the
circumference.
The straight line drawn from A to B shall fall within the
circle.
Construction. — ^Find D the centre of the circle ABC
(III. 1), and join DA, DB.
In AB take any point E ; join DE, and
produce it to the circumference in F.
Proof. — Because DA is equal to DB,
the angle DAB is equal to the angle
DBA (I. 5).
\x \ X / "^^ because AE, a side of the triangle
= z DAB, .^;~ W>a DAE, is produced to B, the exterior angle
^"deb:* ^ DEB is greater than the interior and
DBE. opposite angle DAE (I. 16).
But the angle DAE was proved to be equal to the angle
DBE;
Therefore the angle DEB is also greater than DBE.
But the greater angle is subtended by the greater ei4e
(L 19);
;vs»v& Tiefefore D5 is grcsAe? ^Sft»»,T)lK
PROPOsiTioi^s. 257
But PB is equal to DF; therefore DF is greater
than DE, ani the i)oint E is therefore withm the
circle.
In the same manner it may be proved that every point in
AB lies within the circle.
Therefore the straight line AB lies within the cii'cle.
Therefore, if any two points, &c. Q.E,D,
Proposition 3.— Theorem.
If a straight line drawn through the centre of a circle bisect
a straight line in it which does not pass through the centre, it
shaU cut it at right angles ; and conversely^ if it cut it at right
angles, it shaU bisect it.
Let ABC be a circle, and let CD, a straight line drawn
through the centre, bisect any straight line AB, which does
not pass through the centre.
CD shall cut AB at right angles.
Construction. — ^Take E, the centre of the circle (III. 1),
andjoinEA, EB.
Proof. — Because AF is equal to FB
(Hyp.), and FE common to the two triangles
AFE, BFE, and the base EA equal to the
base EB (I. Def. 15),
Therefore the angle AFE is equal to the
angle BFE (I. 8); \/ I \/ IpE^'I^fd
Therefore each of the angles AFE, BFE ^^^^ZE^^ e^Zun
is a right angle (I. def 10); "^ every ^^
Therefore the straight line CD, drawn through the centre, ^^p^^
bisecting another, AB, that does not pass through the centre,
also cuts it at right angles.
Conversely, let CD cut AB at right angles.
CD shall bisect AB ; that is, AF shall be equal to FB
— ^the same construction being made.
Proof. — Because the radii EA, EB are equal, the angle
EAF is equal to the angle EBF (I. 5),
And the angle AFE is equal to the angle BFE
(Hyp.),
Therefore, in the two triangles EAF, EBF, there are two
angles in' the one equal to two angles in tliQ o\Xx«t^ es5J^\Rk
258
GEOMETRY,
each, and the side EF, which is opposite to one of the equal
angles in each, is common to both;
Therefore their other sides are equal (L 26);
Therefore AF is equal to FB.
Therefore, if a straight line, <kc, Q,E,D,
If they
bisect each
other,
EF bisects
AC at
right
angles.
AndEF
bisects BD
at right
angle^.
.•.ZFEA
(?4FEB.
Proposition 4.—Theorem,
If in a circle two straight lines cut one another, which
do not both pass through tlie centre, they do not bisect ecuHh
otJier,
Let ABCD be a circle, and AC, BD two straight lines in
it, which cut one another at the point E, and do not both
pass through the centre.
AG, BD shall not bisect one another.
If one of the lines pass through the centre, it is plain that
it cannot be bisected by the other, which does not pass through
the centre.
But if neither of them pass through the centre, if possible,
let AE be equal to EC, and BE to ED.
Construction. — Take F, the centre of the circle (III. 1),
and join EF.
Because FE, a straight line drawn
through the centre, bisects another line
AC, which does not pass through the
Ak f ^-^X centre (Hyp.), therefore it cuts it at fight
angles (IIL 3) ;
B>r " ^< / Therefore the angle FEA is a right
angle.
Again, because the straight line FE bisects the straight
line BD, which does not pq3s through the centre (Hyp.),
therefore it cuts it at right angles (III. 3);
Therefore the angle FEB is a right angla
But the angle FEA was shown to be a right angle ;
Therefore the angle FEA is equal to the angle FEB, the
less to the greater, which is impossible;
Therefore AC, BD do not bisect each other/
Therefore, if in ?i circle, &c. Q,E, D,
PHOPOSITIONS, 259
Proposition 5.— Theorem.
If two circles cut one another, tliey shall not have tfie sa/ine
centre.
Let the two circles ABC, CDG cut one another in the
points B, C.
They shall not have the same centre.
For, if it be possible, let E be their centre ; join EC, and
draw any straight line EFG, meeting the ^
circles in F and G.
Proof. — Because E is the centre of the A/
circle ABC, EC is equal to EF (I. Def.
15).
And because E is the centre of the
circle CDG, EC is equal to EG. X^s^ ^^ and =eg
But EC was shown to be equal to EF;
Therefore EF is equal to EG (A3C 1), the less to the greater, /.£?=£€
which is impossible ;
Therefore E is not the centre of the circles ABC, CDG.
Therefore, if two circles, &c, Q,E,D,
Proposition 6.— Theorem.
If one circle tomh another internally, they sIkffM not have
the same centre.
Let the cii-cle CDE touch the circle ABC internally in the
point C.
They shall not have the same centre.
CoNSTBUCTiON. — For, if it be possible, let F be their centre; if they hai
join FC, and draw any straight line FEB, ^ SX"f
meeting the circles in E and B. '
Pboof. — Because F is the centre of
the circle ABC, FC is equal to FB (I Def.
15).
And because F is the centre of the \ \^ / / ^^"iS,
circle CDE, FC is equal to FE. ^ ^^ / ^na^iB.
But FC was shown to be equal to FB ;
Therefore FE is equal to FB (Ax. 1), the less to the greater, .-.fesFB
which is impossible ;
Therefore F is not the centre of the circlea A^C, C!5Yy^.
Therefore, if one circle, &c, Q,E,D.
260 GEOMETRY.
Proposition 7.--Theorem.
If any 'point be taken in the dia/meter of a circle, which is
not tlie centre of all tJie straigM lines which can he drawn
from this point to tlie circurnference, tJie greatest is that tn
which the centre is, and the other part of tlwi diameter is ihe
least ; amd, of any otiiers, that which is nearer to the straight
line which passes through the centre is ahoays greater than
one more remote ; and from the sarnie point there cam he dra/um
to the circumference two straight lines, and only ttvo, which
are equal to one another, one on each side of the s/wrtest line.
Let ABCD be a circle, AD its diameter, and E its
centre, in which let any point F be taken which is not the
centre.
FA>FB Of all the straight lines FB, FC, FG, &c., that can be
^^^:?FD ^rawn from F to the circumference, FA, which passes
tho'ieast. through the centre, shall be the greatest ;
FD, the other part of the diameter AD, shall be the least;
And of the others, FB, the nearer to FA, shall be greater
than FC, the more remote ; and FC greater than FG.
j,^^ Construction. — Join BE, CE, GE.
BE+EF Proof. — Because any two sides of a triangle are greater
crAr>BF ^j^ than the third side, BE, EF are greater
than BF (I. 20).
But AE is equal to BE; therefore
AE, EF, that is, AF is greater than BF.
Again, because BE is equal to CE,
and EF common to the two tnangles BEF,
CEF, the two sides BE, EF are equal to
the two CE, EF, each to each.
But the angle BEF is greater than the angle CEF ;
And Therefore the base FB is greater than the base FC
>-baseFO ^ '*
In the same manner it may be shown that FC is greater
than FG.
Ajrain Again, because GF, FE are greater than EG (I. 20),
>EG^^ and that EG is equal to ED ;
aiia.-.>ED. Therefore GF, FE are greater than ED.
/.OF^FD Take away the common ^ort FE, and the remainder GF
is greater thau the xeixxsajx^ev ^T> ^^Xtu ^"^^
PROPOSITIONS. 261
Therefore FA is the greatest, and FD the least, of all the
straight lines from F to the circuiiiference ; and FB is greater
than FC, and FC than FG.
Also, there cannot be drawn more than two equal straight
Jiines from the point F to the circumference, one on each side
of the shortest line.
Construction. — ^At the point E, in the straight line EF,
make the angle FEH equal to the angle FEG (I. 23), and
joinFH.
Proof. — Because EG is equal to EH, and EF common to ^*"8^^
the two triangles GEF, HEF, the two sides EG, EF are and hep
equal to the two sides EH, EF, each to each ; fn^e^^
And the angle GEF is equal to the angle HEF (Const.); respect.
Therefore the base FG is equal to the base FH (I. 4).
But, besides FH, no other straight line can be drawn from
F to the circumference equal to FG.
For, if it be possible, let FK be equal to FG ;
Then, because FK is equal to FG, and FH is also equal to ^^P
FG, therefore FH is equal to FK; lOso^Fi
That is, a line nearer to that which passes through the J^^Sl,!,
centre is equal to a line which is more remote; which is im-
possible by what has been already shown.
Therefore, if any point, &c. Q,E,D,
Proposition 8.— Theorem.
If any point he taken without a circle, and straight Urns he
drawn from it to the circu/m/erence, one of which passes
Uwimgh the cenVre ; of tliose which fall on t/ie concave circum-
ference, tlie greatest is tJiat which passes through the centre,
and oftJie rest, that which is nearer to tlie one passing through
t/ie centre is always greater tJian one more remote ; hut of ilvose
which faU on tlie convex circun\ference, the least is that he-
tween the point without the circle and the diameter ; and of
the rest, that which is nearer to the least is always less than
one more remx>te ; and from the sams point there can he drawn
to the circumference two straight lines, and only two, which
are eqvxd to one anotlier, one on each side of the least line.
Let ABC be a circle, and D any point without it, and
from D let the straight linea DA, DE, D^, DO \i^ ^;x«w\3k
363
to the mnnmiiieTeBce, of whidi DA paaes ihroo^ tlw
centre.
I>A>DE Of those which fall on the craicKTe part of the circiim-
^{^ ference AEFC, the greatest shall he DA, which j»nea
throng the centre, and the nearer to it shall be greato'.
than the more remote, viz., D£ greater than DF, ud D¥
greater than DC,
ind Bnt of those which &11 on the convex circcinferaice GEIJS,
^^^ the least shall be DO between the point D and the diameter
■OB AG, and the nearer to it shall be lem than the more remote,
yiz., DK less than DL, and DL less than DH.
CoxsTBUcnos.— rTake M, the centre ttf the circle ABC
(m 1), and join ME, MP, MC, Mff, ML, M K .
Pboof. — Because any two ndea of a
triangle are greater Htsta the tiurd mde,
" EM, MD are greater than HD (L 20).
Bob EM is equal to AM ; therefwe
AM, MD are greater than ED — ^that
is, AD is greater than ED.
Again, becanse EM ia eqnal to FM,
and MD common to the tWo triangles
EMD, FMD ; the two bides EM, MD
are equal to the two sides FM, MD,
each to each;
But the angle KMD is greater i3»m
the angle EMD ;
Thea«forethe base ED is greater than
the base FD (L 24).
In lite manner it maj be shown that
PD is greater than CD;
Therefore DA ia the greatest, and DE greater than i>S
and DP greater than DC.
Again, because ME, ED are greater than MD (L 20),
and MK ia equal to MG;
The remainder KD is greater than the remaindei' GD—
that is, GD ia less than KD.
And because MK, DK are irnvn to the point K within
the triangle MLD from M and D, the extremities of its side
MD;
Therefore MK,DK. axft\esaXV«&^\.,liD <I. 21).
PROPOSITIONS. 263
But MK is equal to ML ; therefor^ the remainder KD is
less than the remainder LD.
In like manner it may be shown that LD is less than HD.
Therefore DG is the least, and KD less than DL, and DL
less than DH.
Also, there can be drawn only two equal straight lines from
the point D to the circumference, one on each side of the
least line.
Construction. — At the point M, in the straight line MD,
make th& angle DMB equal to the angle DMK (L 23), and
join DB;
Proof. — Because MK is equal to MB, and MD common Triangles
to the two triangles KMD, BMD; the two sides KM, MD andSMD
are equal to the two sides BM, MD, each to each; f^?2"*^
And the angle DMK is equal to the angle DMB (Const); nspwu
Therefore the base DK is equal to the base DB (I. 4).
But, besides DB, no other straight line can be drawn from
D to lie circumference equal to DK.
For, if it be possible, let DN be equal to DK.
Then, because DN is equal to DK, and DB is also equal DN=i)K
to DK, therefore DB is equal to DBT (Ax. 1); =Dfe'.
That is, a line nearer to the least is equal to one which is J^l» !■
more remote; which is impossible by. what has been already ^**^^
shown.
Therefore, if any point, &c. Q.EiD.
Proposition 9.— Theorem*
if a point he taken within a circle^ from which there tan he
drcuum more than two equal straight lines to tlie circumference ^
that point is tlie centre of the circle.
Let the point D be taken within the circle ABC, from
which to the circumference there can be
drawn more than two equal straight lines,
viz., DA, DB, DC.
The point D shall be the centre of the r[
circle.
Construction. — For if not, let E be
the centre ; join DE, and produce it to
the circwaafereuce in F and G,
264 GEOMETRY.
Proof. — ^ThenFGis adiameterof the circle ABC (T.De£ 17).
If D be not And because in FG, the diameter of the circle ABC, there
the centre. ^ ^^^^ ^^ ^^^ jy^ ^^^ ^^ ^^^^ .
DG^DC Therefore DG is the greatest straight line from D to the
>DA. circumference, and DC is greater than DB, and DB greater
than DA (m. 7);
But they But theso lines are likewise equal, by hypothesis; which is
'^^ impossible.
Therefore E is not the centre of the circle ABC.
In like manner it may be demonstrated that any other
point than D is not the centre ;
Therefore D is the centre of the circle ABC.
Therefore, if a point, &c. Q.E.J),
PropoBition 10. — ^Theorem.
One circun^erence of a circle cannot cut anotiter in more
than two points.
Construction. — ^If it be possible, let the circumference
If possible, _j^^ ABC cut the cirtnimference DEF in
more than two points, viz., in the points
B,G,F.
Take the centre K of the circle ABC
(III. 1), and join ElB, KG, EJF.
Proof. — Because K is the centre of
the circle ABC, the radii KB, KG, KF
are all equal.
the two And because within the circle DEP there is taken the
havc'^the point K, from ■ which to the circumference DEF fall more
8ame
centre.
than two equal straight lines KB, KG, KF, therefore K is
the centre of the circle DEF (III. 9).
But K is also the centre of the circle ABC (Const.) ;
Therefore the same point is the centre of two circles which
cut one another; which is impossible (III. 5).
Therefore, one circumference, &c. Q.E.D.
Proposition 11.— Theorem.
If one circle touch another internally in any point, the
straight line which joins tlieir centres, being produced, shall
pass throv/gh tlmt point.
Let the circle ADEi \youda. \5aft ca<^^ KSS^Sss^rasillY in the
PROPOSITIONS.
265
point A; and let F be the centre of the circle ABC, and G
the centre of Ae circle ADE.
The straight line which joins their centres, being produced,
shall pass through the point of contact A
Construction. — For, if not, let it pass otherwise, if if not,
possible, as FGDH. Join AF and AG.
Proof. — Because AG, GF are greater
than AF (I. 20), and AF is equal to HF ^ ,
(I. def. 15); ^
Therefore AG, GF are greater than HF.
Take away the conunon part GF, and the
remainder AG is greater than the remainder
HG.
But AG is equal to DG (I. Def. 15) ; But ag
Therefore DG is greater than HG, the less than the f g^;^
greater; which is impossible. ho.
Therefore the straight line which joins the centres, being
produced, cannot fall otherwise than upon the point A, that
IS, it must pass through it.
Therefore, if one circle, &c. Q.E,D.
AG>HG.
Proposition 12. — Theorem.
If two circles touch each other eodernally in any point j the
straight line which joins their centres slmtl pass through that
point
Let the two circles ABC, ADE touch each other externally
in- the point A; and let F be the centre of the cncle ABC,
and G the centre of the circle ADE;
The straight line which joins
their centres shall pass through
the point of contact A.
Construction. — For, if not,
let it pass otherwise, if possible,
as FCDG. Join FA and AG.
Proof. — Because F is the
centre of the circle ABC, FA
equal to FC (I. Def. 15).
IS
And because G is the centre of the circle ADE, GA is
equal to GD;
u not,
S66
OEOMETttlf.
FGfs>
FA+AG,
but it is
oXaoless,
Therefore FA, AG ai^e equal to FC, DG (Ax. 2).
Therefore the whole FG is greater than FA, 'AG.
But FG is also less than FA, AG (L 20) ; which is im-
possible.
Therefore the straight line which joins the centres of the
circles shall not pass otherwise than through the point A,
that is, it must pass through it.
Therefore, if two circles, &c. Q,E,D.
Proposition 13.— Theorem.
One circle cannot touch anotlier in more points than one,
whether it toiLch it internally or extemaUy.
I. First, let the circle EBF touch the circle ABC internally
in the point B.
Then EBF cannot touch ABC in any other point.
If possible, Construction. — If it be possible, let EBF touch ABC in
i**Dk£S^^ another point D; join BD, and draw GH bisecting BD at
right angles (L 10, 11).
Proof. — Because the two points B, t) are in the circum-
ference of each of the circles, the straight line BD falls within
each of them (III. 2).
Therefore th^ centre of each fcircle is in the straight line
(jtH, which bisects BD at right angles (III. 1 cor.)
ti,en Therefore GH passes through the {)oint of contact (III. 11).
GHpa^^ But GH does not pass thrdugh the point of contact,
the point becausc the points B, D are but of the line of GH; which
ofcontact, ;« oh<?nrfl
which it ^^ aosura.
does not. Therefore one circle cannot touch another internally in
more points than one.
t>R0P0SITI01fS. 267
11 Next, let the circle ACK touch the circle ABC
externally in the point A.
Then ACK cannot touch ABC in any other poiht.
Construction. — If it be possible, let ACK Miuch ABC in u poesibie
another point C. Join AC. ^ — ^ to c iSwf
Proof. — Because the points A, C are
in the circumference of the circle ACK, the
straight line AC must fall within the circle
ACK (IIL 2).
But the circle ACK is without the circle
ABC (Hyp.);
Therefore the straight line AC is without I J then
the circle ABC. \ / tShSSt
But because the two points A, C are in ^<^ / abc* '^**
the circumference of the circle ABC, the whiciiis
straight Hue AC falls within the circle ABC (III 2); which »^««*-
is absurd.
Therefore one circle cannot touch another elctemally in
more points than one.
And it has been shown that one circle cannot touch an-
other internally in more points than one*
Therefore, one circle, &c. Q,E,Di
Proposition 14. — Thebreni.
Equal straight lines in a dfcte are equally dista/nt/rom the
cervl/re; andy conversely y tliose which are equally distant from
the centre twe equal to one another.
Let the straight lines AB, CD, in the cii'cle ABDC, be
equal to one another. «-
Then they shall be equally distant from the bentrti;
CoNSTRUCTiONi — ^ake tl, the centre of the circle ABDC
(IIL 1).
Prom E draw EF, EG, perpendiculars to AB, CD (I. 12).
Join EA, EC.
Proo^. — Because the Bti-aight line EF, passing through
the centre, cuts the straight line AB, which does not pass
through the centre, at right angles, it also bisectsr it (III. 3).
Therefore AF is equal to FB, and AB ia AoM\Aa cii KS .
For the like reason CD is double of CG.
268 GEOMETRY
AF = CO, But AB is equal to CD (Hyp.) ; therefore AF is equal to
CG (Ax. 7).
And because AE is equal to CE, the square on AE is
equal to the square on CE.
But the squares on AF, FE are equal to
the square on AE, because the angle AF£
is a right angle (I. 47).
For the like reason the squares on CG,
GE are equal to the square on CE ;
AP2+FES Vf y Therefore the squares on AF, FE
= CG2 + ^v y^ are equal to the squares on CG, GE
But the square on AF is equal to the square on CG,
because AF is equal to CG;
Therefore the remaining square on FE is equal to the
remaining square on GE (Ax. 3) ;
.•.EF=EG. And therefore the straight line EF is equal to the straight
line EG.
But straight lines in a circle are said to be equally distant
from the centre, when the perpendiculars drawn to them
from the centre are equal (III. Def. 4) ;
Therefore AB, CD are equally distant from the centre.
Conversely, let the straight lines AB, CD be equally dis-
tant from the centre, that is, let EF be equal to EG;
Here Then AB shall be equal to CD.
and"^ ' Proof. — ^The same construction being made, it may be
APi+EFs demonstrated, as before, that AB is double AF, and CD
EG-'. double of CG, and that the squares on EF, FA are equal to
the squares on EG, GC.
But the square on EF is equal to the square on EG,
because EF is equal to EG (Hyp.);
Therefore the remaining square on FA is equal to the
remaining square on GC (Ax. 3),
.-. AF = And therefore the straight line AF is equal to the straight
CG» ^' line CG.
But AB was shown to be double of AF, and CD double
ofCG;
Therefore AB is equal to CD (Ax. 6);
Therefore, equal straight lines, <fec. Q,E,D.
PROPOSITIONS. 269
Proposition 15. — Theorem.
TJie diameter is the greatest straight line in a circle; and,
of all others, that which is nearer to tlie centre is always greater
tlian one more remote; and tlie greater is nearer to tlie centre
iJian the less.
Let ABCD be a circle, of which AD is a diameter, and E
the centre ; and let BC be nearer to the centre than FG.
Then AD shall be gi*eater than any straight line CB,
which is not a diameter; and BC shall be greater than FG.
Construction. — From the centre E draw EH EK per-
pendiculars to BC, FG (I. 12), and join
EB, EC, EF.
Proof. — Because AE is equal to BE,
and ED to EC,
Therefore AD is equal to BE, EC. PI , »K \ ] or^ "^ ^^
But BE, EC are greater than BC \ \\ / ad>bc,
(1.20);
Therefore also AD is greater than BC.
And because BC is nearer to the centre than FG and
(Hyp.), EH is less than EK (III. Def. 5). eh<ek.
But, as was demonstrated in the preceding proposition,
BC is double of BH, and FG double of FK, and the squares eh«+ hb^
on EH, HB are equal to the squares on EK, KF. ^^^'^ +
But the square on EH is less than the square on EK,
because EH is less than EK ;
Therefore the square on HB is greater than the square on
KF, and the straight line BH greater than FK; hb>fk.
And therefore BC is greater than FG.
Conversely, let BC be greater than FG.
Then BC shall be nearer to the centre than FG, that is,
the same construction being made, EH shall be less than EK.
Proof. — Because BC is greater than FG, BH is greater
thanFK.
But the squares on BH, HE are equal to the squares on
FK, KE;
And the square on BH is greater than the square on FK,
because BH is greater than FK;
Therefore the square on HE is less than the sci^vaxoi orcv
KE, and the stnight line EH less tliaA^^\
270
GEOMETRY,
And therefore BC is nearer to the centre than FG (ifl.
def. 5).
Therefore, the diameter, &c. Q.E,D.
Take any
point F in
AE,
then
DF>DA
and . *. i:^
DC.
DrawDH
at right
aniiries to
HO, then
DH <= DA,
and .*. •<
DK.
greater than DA
Proposition 16.r-Theorem.
TJie straight line drawn at right angles to tJie diameter of a
circle, from the extremity of it, falls without the circle; and a
straight line, making an a>cute angle with the diameter at its
extremity, cuts t/ie circle.
Let ABC be a circle, of which D is the centre, and AB a
diameter, and AE a line, drawn from A perpendicular to
AB.
The straight line AE shall fall without the circle.
Construction. — In AE take any point F; join DF, and
let DF meet the circle in C.
Proof. — Because DAF is a right angle,
it is greater than the angle AFD (I. 17);
Therefore DF is
(I. 19).
But DA is equal to DC; therefore DF
is greater than DC.
Therefore the point F is without the
circle.
In the same manner it may be shown that any other point
in AE, except the point A, is without the circle.
Therefore AE falls without the circle.
Again, let AG make with the diameter the angle DAG
less than a right angle.
The line AG shall fall within the circle, and hence cut it.
Construction. — ^From D draw DH at
light angles to AG, and meeting the cir-
cumference in K (I. 12).
Proof. — Because DHA is a right angle,
and DAH less than a right angle;
Therefore the side DH is less than the
side DA (I. 19).
But DK is equal to DA; therefore DH
la leaa tlaajv "DK.
jTh^refore the point "& i^ ^^m^Dci^ ccc^^.
PaOPOSITIONS,
271
Therefore the straight line AG cuts the circle.
Therefore, the straight line, &c. Q,E,D,
Corollary. — From this it is manifest that the straight
line which is drawn at right angles to the diameter of a
circle, from the extremity of it, touches the circle (III. Def.
2) ; and that it touches it only in one point, because if it did
meet the circle in two points it would fall within it (III. 2).
Also it is evident that there can be but one
which touches the circle in the same point.
straight line
Proposition 17.— Problem.
To draw a straight line from a given pointy either without
or in t/ie circumference, which iJiall touch a given circle.
First, let the given point A be without the given circle
BCD.
It is required to draw from A a straight line which shall
touch the given circle.
Construction. — Find the centre E of the circle (III. 1),
and join AE.
From the centre E, at tlie distance
EA, describe the circle AFC
From the point D draw DF at right
angles to EA (I. 11), and join EBF
and AB.
TJien AB sJuill touch tli/e circle BCD,
Proof. — ^Because E is the centre of
the circles AFG, BCD, EA is equal to
EF, and ED to EB;
EA,EDr
spectivel;
Therefore the two sides AE, EB are equal to the two sides ^d / e
common
FE, ED, each to each;
And the angle at E is common to the two triangles AEB,
FED;
Therefore the base AB is equal to the base FD,^ and the
triangle AEB to the triangle FED, and the other angles to
the other angles, each to each, to which the equal sides are
opposite (I. 4);
Therefore the angle ABE is equal to the ang^le ¥I>12i,
5ut the qjogle FDE is a right angle (Coixat.y,
.'. Z AE
272 GEOMETRY.
Therefore the angle ABE is a right angle (Ax. 1),
And EB is drawn from the centre (Const.)
But the straight line drawn at right angles to a diameter
of a circle, from the extremity of it, touches the circle (III,
16, cor.);
.'. AB Therefore AB touches the circle, and it is drawn from the
lh"ciS.. given point A.
Next, let the given point be in the circumference of the
circle, at the point D.
Draw DE to the centre E, and DF at right angles to DE;
Then DF touches the circle (III. 16, cor.)
Therefore, from the given points A and D, straight lines,
AB and DF, have been drawn, touching the given ciixjle
BCD. Q,E.F.
Proposition 18. — Theorem.
If a straight line touch a circle, tlie straight line drawn
from tlie centre to the point of contact shall be perpendicular
to the line touching the circle.
Let the straight line DE touch the circle ABC in the
point C; take the centre F (III. 1), and draw the straight
line FC.
FC shall be perpendicular to DE.
ifnot.8ui)- Construction. — ^For, if not, let FG be dsrawn from the
peniendi- ^ point F perpendicular to DE, meeting
^"lar. /"^ ^\ the circumference in B.
Proof. — Because FGC is a right angle
F ^ (Hyp.), FCG is an acute angle (I. 17),
and to the greater angle the gi*eater side
is opposite (I. 19);
Therefore FC is greater than FG.
Then must ^ ^ ^^^ ^^* ^^ ^ ^^^ ^ ^^ > therefore FB
rii>FG. is greater than FG; the part greater than the whole, which
is impossible.
Therefore FG is not perpendicular to DE.
In the same manner it may be shown that no other straight
line from F is perpendicular to DE, but FC; therefore FC
13 perpendicular to DE.
Therefore, if a Btrai^\i^/\m^> k^v Q,.E,p.
PROPOSITIONS.
273
Proposition 19.— Theorem.
If a straight line touch a circle, and from the point of con-
tact a straight line he dravm at right angles to the touching
line, the centre of the circle shall he in that line.
Let the straight line DE touch the circle ABC in C, and
from C let CA be drawn at right angles to DE.
The centre of the circle shall be in CA.
Construction. — ^For, if not, if possible, let E be the centre,
and join CF.
Proof. — Because DE touches the circle
ABC, and EC is drawn from the assumed
centre to the point of contact.
Therefore EC is perpendicular to DE
(III. 18);
Therefore ECE is a right angle. ^ g U
But the angle ACE is also a right angle (Const);
Therefore the angle ECE is equal to the angle ACE; the
less to the greater, which is impossible.
Therefore F is not the centre of the circle ABC.
In the same manner it may be shown that no other point
which is not in CA is the centre; therefore the centre is in
CA.
Therefore, if a straight line, &c. Q,KD.
Proposition 20. — Theorem.
The a/ngle at the centre of a circle is double of tli,e angle at
the circumference, upon the same hose, that is, upon the same
a/rc.
Let ABC be a circle, and BEC an angle
at the centre, and BAC an angle at the cir-
cumference, which have the same arc BC
for their base.
The angle BEC shall be double of the
angle BAC.
Case I. — First, let the centre E of the
circle be within the angle BAC.
CoNSTEucTiQ^.-r-Jpiu AE, and produce it to the circmu*
^enceiQFt
B 9
If not.
take F the
centre, out
of the line,
Then
ZACE =
ZFCE,
beins; right
anglds.
274
GEOMETRY.
l PEC=
2z EAC,
and ZFEr
B2/CAB.
/.taklngr
Proof. — Because EA is equal to EB, the angle EAB is
equal to the angle EBA (I. 5) ;
Therefore the angles EAB, EBA are double of the angle
EAB.
^ BEF = But the angle BEF is equal to the angles EAB, EBA
Z EAB + /y Q0\ . ^ ®
Z EBA \}-* ^^)f
^2 zEAB, Therefore the angle BEF is double of the angle EAB.
Im^^ For the same reason the angle FEC is double of the angle
2/BAC. EJ,0.
.•.zBEO Therefore the whole angle BEC is double of the whole
•=»^»^^- angle BAG.
Case IL--^Nexty let the centre E of the
circle be without the angle BAG.
GoNSTRUCTioN. — Join AE, and produce it
to meet the circumference in F.
Proof. — It may be demonstrated, as in
the first case, that the angle FEG is double
of the angle FAG, and that FEB, a part of
FEG, is double of FAB, a part of FAG;
Therefore the remaining angle BEG is double of the re-
the differ- mainiuff anffle BAG.
z BEC = Therefore, the angle at the centre, d^. Q.KD,
2 4. BAG,,
Proposition SI.— Theorem.
TJie angles in the sa/im eegment of a circle are equal to cm
another.
Let ABGD be a circle, and BAD, BED angles in the same
segment BAED.
The angles BAD, BED shall be equal to one another.
Gase L — First, let the s^ment BAED
be greater than a semicircle.
CoKSTRtJCTiON. — ^Take P, the centre of the
circle ABGD (in. 1), and join BP, DF.
Proof. — Because the angle BFD is at iJie
centre, and the angle BAD at the circum-
ference, and that they hare the same ato for
their base, namely, BOD;
zBFD=a Therefore the angle BFD is double of the angle BAD
^^iUD, ^jjj 20).
PIJOPOSITIONS.
275
For the same reason, the angle BFD is double of the angle and
also =
2 I BED.
.*. z BAD
= I BED.
BBDj
Therefore the angle BAD is equal to the
aogle BED (Ax. 7).
Case II. — ^Next, let the segment BAED
be not greater than a semicircle.
CoNSTBUCTiON. — Draw AF to the centre,
and produce it to C, and join CK
Pboof. — ^Then the segment BADC is
greater than a semicircle, and therefore the angles BAG,
BEG in it are equal by the first case.
For the same reason, because the segment GBED is greater and
than a semicircle, the angles GAD, CED are equal. j ^^ =
Therefore the whole angle BAD is equal to the whole .-. z. bad
angle BED (Ax, 2), =^ bed.
Therefore, the angles in the same segment, <&c. Q,E,D,
/. BAC =
/ BEG,
Proposition SS.— ThecHrem.
The opposite angles of cmy quadrilateral figwire inscribed in
a circle are togdher equal to two right ambles.
Let ABCD t)e a quadrilateral figui^ inscribed in the circle
ABGD.
Anv two of its opposite angles shall be together equal to
two right angles.
Construction. — Join AC, BD.
Pboop. — ^Becaiiito the three angles of
every triangle are together equal to two
right angles (I. 32),
The three angles of the triangle CAB,
namely, CAB, ACB, ABC, are together
equal to two right angles.
But the fthgle CAB is equal to the angle
CDB, because they are in the same seg-
ment CDAB (IIL 21);
Ahd thd Ikiigie ACB \a equal td the angle ADB, because
tliey are Jit the ^ame segment ADCB;
Therefore the two angles CAB, ACB are together eo^JAlta
the i«rhole angle ADO (Ax 2),
Z CAB +
Z ACB +
Z ABC =
2 right
angles.
The first
two toge-
ther =
ZCDB +
Z ADB =
Z ADC.
276 GEOMETBY.
To each of these equals add the angle ABC;
Therefore the three angles CAB, ACB, ABC are equal to
the two angles ABC, ADC.
But the angles CAB, ACB, ABC are together equal to
two right angles (I. 32);
.*. z ADC Therefore also the angles ABC, ADC are together equal
i 2^rigM' *o *wo right angles.
Angles. In like manner it may be shown that the angles BAD,
BCD are together equal to two right angles.
Therefore, the opposite angles, &c. Q.E.D,
Proposition 23.— Theorem.
Upon the same straight liney and on the same side of it, then
cannot he two similwr segments of circles not coinciding vnth
one another.
If possible. If it be possible, on the same straight line AB, and on
the same side of it, let there be two similar segments of
circles ACB, ADB not coinciding with
one another.
Construction. — ^Then, because the circle
ACB cuts the circle ADB in the two points
A, B, thej cannot cut one another in any
other point (III. 10);
Therefore one of the segments must M
within the other.
Let ACB fall within ADB; draw the straight line BCD,
and join AC, AD.
Proof. — ^Because the segment ACB is similar to the s^-
ment ADB (Hyp.), and that similar segments of circles
contain equal angles (IIL Def. 11);
exterior^ Therefore the angle ACB is equal to the angle
interior" ADB; that is, the exterior angle of the triangle ACD,
Jg?<*ppo- equal to the interior and opposite angle; which is impoe-
ZADC. sible (1. 16).
Therefore, there cannot be two similar segments of cirdes
on the same straight line, au4 9Xi the same s^de pf it^ wMdi
dp not coincide, Q,JS,D,
PROPOSITIONS.
277
Proposition 24. — Theorem.
Simila/r segments of circles vpon equal straight lines me
egtml to one another.
Let AEB, CFD be similar segments of circles upon the
equal straight lines AB, CD.
The segment AEB shall
be equal to the segment
CFD.
Proof. — ^For if the seg-t
ment AEB be applied to the
segment CFD, so that the point A may be on the point C,
and the straight line AB on the straight line CD,
Then the point B shall coincide wiUi the point D, because
AB is equal to CD.
And ^e straight line AB coinciding with CD, the seg-
ment AEB must coincide with the segment CFD (III. 23);
and is therefore equal to it.
Therefore, similar segments, &c. Q.E,D,
Proposition 25. — Problem.
A segment of a circle being given, to describe the ctrde of
which it is the segmerU.
Let ABC be the given segment of a circle.
It is required to describe the circle of which ABC is a
segment.
Construction. — Bisect AC in D (I. 10).
From the point D draw DB at right angles to AC (1. 11),
and join AB.
They are
equal,
because
they must
n coincide by
Prop. 28.
Case I. — First, let the angles ABD, BAD be equal to
one another.
I%en D shall he the centre of the circle requxred.
278
GEOMEXEY.
When
ZABD =
DA = DB
and
.*. Dthe
centre.
Hake
z BAE
i^ ABD.
EA =
£B
andEA =
EC.
.-. EA =
EB = EC,
and there-
fore E is
the centre.
Proof. — Because the angle A£D is equal to the angle
BAD (Hyp.);
Therefore DB is equal to DA (L 6).
But DA is equal to DC (Const);
Therefore DB is equal to DC (Ai. 1).
Therefore the three straight lines DA, DB, DC aie iU
equal;
And therefore D is the centre of the circle (llL 9).
Hence, if from the centre D, at the distance of any of the
three lines, DA, DB, DC a circle be described, it will pass
through the other two points, and be the circle required.
Case IL — Next, let the angles ABD, BAD be not equal
to one another.
CoNSTEUCTiON. — At the point A, in the straight line AB,
make the angle BAE equal to the angle ABD (L 23);
Produce BD, if necessary, to E, and join EC.
Then E ahaU he the centre of the circle required.
Proof. — Because the angle BAE is equal to the angle
ABE (Const.), EA is equal to EB (I. 6).
And because AD is equal to CD (Const), and DE is com-
mon to the two triangles ADE, CDE,
The two sides AD, DE are equal to the two sides CD,
DE, each to each;
And the angle ADE is equal to the angle CDE, for each
of them is a right angle (Const.);
Therefore the base EA is equal to the base EC (I. 4).
But EA was shown to be equal to EB;
Therefore EB is equal to EC (Ax. 1).
Therefore the three straight lines EA, EB, EC are all
equal;
And therefore E is the centre of the circle (IH. 9).
Hence, if from the centre E, at the distance of any of the
three lines EA, EB, EC, a circle be described, it will pass
through the other two points, and be the circle required.
In ^e first case, it is evident that, because the centre D
is in AC, the segment ABC is a semicircle.
In the second case, if the angle ABD be greater than
BAD, the centre E falls without &e segment ABC, which is
therefore less than a semicircle;
But if the angle ABD \)^ \^i8i& ^i^t^aai "^^ «S!^<^ BAD, the
PROPOSITIONS. 279
centre E £ei11s within the segment ABC, which is therefore
greater than a semicircle.
Therefore, a segment of a circle being given, the circle has
been described of which it is a segment. Q,U,F.
Proposition 26.— Theorem.
In equal circles, equal cmglea stamd tspon eqtudarcSf wlietfier
they he a/t the centres or at the circumferences*
Let ABO, DEF be equal circles, having the equal angles
BGC, EHP at their centres, and BAG, EDF at their cir-
cumferencea
The arc BKC shaU be equal to the arc ELF.
Construction. — Join BO, EF.
Proof. — Because the circles ABC, DEF are equal (Hyp.),
the straight lines from their centres are equal (IIL def. 1);
Therefore the two sides BG, GC are equal to the two sides Triangleo
EH, HP, each to each; |gg^
And the angle at G is equal to the angle at H ^Hyp.) ; equal in
Therefore the base BC is equal to the base EF (I. 4). I^f.'*"
And because the angle at A is equal to the angle at D
(Hyp.),
c z
The segment BAC is similar to the segment EDF (III. .*. seg-
^«f 11\ mentsBAO
aei. 11), and EDF
And they are on equal straight lines BC, EF. aresimiiar
But simUar segments of circles on equal straight lines are ^ua?^
equal to one another (IIL 24); SJS^^*
Therefore the segment BAC is equal to the segment EDF. .*. are
But the whole circle ABC is equal to the whole circle *^"*^
PEP (Hyp.);
* Therefore the remaining segment BKC ia ^C3^id^ \a ^<!^
remainiTigr Begment ELF (Ax. 3).
280 GEOMETBT.
BKC*i= Therefore the arc BKC is equal to the arc ELF.
arc ELF. Therefore, in equal circles, &c. Q,E,D,
Proposition 27. — Theorem.
In equal circles, the angles which stamd vjpon equal arcs are
equal to one amother, whether they be at the centres or ai the
circwrr^erences.
Let ABO, DEF be equal circles, and let the angles BGO,
EHF, at their centres, and the angles BAG, EDF, at their
circumferences, stand on equal arcs BO, EF.
The angle BGC shall be equal to the angle EHF, and the
angle BAG equal to the angle EDF.
Construction. — If the angle BGG be equal to the angle
EHF, it is manifest that the angle BAG is also equal to tibe
angle EDF (HI. 20, ax. 7).
But, if not, one of them must be the greater. Let BGC
If one z is be the greater, and at the point G, in the straight line
gSiIhe SC^, make the angle BGK equal to the angle EHF
other, the (L 23).
sponding Proop. — ^Because the angle BGK is equal to the angle
^^ ^ter FHF, and that in equal circles equal angles stand on equal
arcs, when they are at the centres (IH. 26) ;
Tierefore the arc BK is equal to the arc EF.
But the arc EF is equal to the arc BC (Hyp.);
Therefore the arc BK is equal to the arc BG (Ax. 1); the
less to the greater, which is impossible.
Therefore the angle BGG is not unequal to the angle
EHF; that is, it is equal to it.
And the angle at A. \e \\fiA£ q£ the angle BGG, and the
angle at D is kali o? tti^ aa^<&lSSS I^Sl.^^-,
PR0]?0SITI0NS. 281
Therefore the angle at A is equal to the angle at D
(Ax. 7).
Therefore, in equal circles, &c. Q,E,D,
Proposition 28. — Theorem.
In equal circles^ equal chords cut off equal a/rcs^ t/ie greater
equal to the greater, and the less equal to the less.
Let ABC, DEF be equal circles, and BC, EF equal chords
in them, which cut off the two greater arcs BAG, EDF, and
the two less arcs BGC, EHF.
The grealer arc BAG shall be equal to the greater
arc EDF, and the less arc BGG equal to the less arc
EHF.
Construction. — ^Take K, L, the centres of the circles Take k
(III. 1), and join BK, KG, EL. LF. ^^t^
Proof. — Because the circles ABC, DEF are equal, their
radii are equal (IIL def. 1).
Therefore the two sides BK, KC are equal to the two
sides EL, LF, each to each;
And the base BC is equal to the base EF (Hyp.) ;
Therefore the angle BKC is equal to the ancle Triangles
Tjix T? /T ft\ ^^ BKC and
JliljJ? (1. ©;. ^ ELF are
But in equal circles equal ancles stand on equal arcs, equal in
when they are at the centres (IIL 26) ; spect
Therefore the arc BGC is equal to the arc EHR
But the whole circle ABC is equal to the whole circle
DEF (Hyp.);
Therefore the remaining arc BAG is equal to the remain-
ing arc EDF (Ax. 3).
Therefore, in equal circles, (fee. Q»jB.D.
282 OEOitETRY.
Proposition 29.— Theorem.
In equal circies eqitcU arcs are tubtended by eqva,l chords.
Let ABC, DEF be equal circles, and let BGC, EHF be
equal arcs in them, and join BC, EF.
The chord BC shall be equal to the chord EF.
ThkeK CoNSTEUCTiON. — ^Take K, L, the centres of the cirdes
^^ (HI. 1), and join BK, KC, EL, LF.
Then pROOF. — Becausc the arc BGrC is equal to titie aro
zeS?^ EHF (Hyp.), the angle BKC is equal to the angle ELP
• (HL 27).
And because the circles ABC, DEF are equal (Hyp.),
their radii are equal (IIL def. 1).
Therefore the two sides BK, KC are equal to the two
sides EL, LF, each to each; and they contain equal angles;
and 80 base Therefore the base BC is equal to the base EF (L 4).
BC = base Therefore, in equal circles, &c. Q»E.D»
Proposition 30.— Problem.
I To bisect a gieen arc, thai is^ to divide it into two equal
parts.
, Let ADB be the given aro.
It is required to bisect it.
Construction. — Join AB, and bisect it in C (I. 10).
I From the point C draw CD at right angles to AB (I. 11),
and join AD and DB.
Then the arc AfiD shall be bisected in tJte
^^^^^^ point I),
inthetri- /^^^^^^^^ Proof. — ^Because AC is equal to CB
mb^mado f^ — J ^^ (Const.), and CD is common to the two tri-
^Bo^ositioifg. ^8J
The two sides AO^ CD are equal to the two sides BO, CD,
each to each ;
And the angle AOD is equal to the angle BCD, because
* each of them is a right angle (Const.) ;
Therefore the base AD is equal to the base BD (I. 4). ^ U;
But equal chords cut off equal arcs, the greater equal to
the greater, and the less equal to the less (IIL 28);
And each of the arcs AD, DB is less than a semicircle,
because DC, if produced, is a diameter (IIL 1, cor.) ;
Therefore the arc AD is equal to the arc DB.
Therefore^ the given arc is bisected in D. Q,J^.F,
Proposition 31.— Theorem.
»
In a circle^ the angle in a semicircle is a right cmgle; hut
the angle in a segment greater tha/n a semicircle is less thorn a
right angle; and tlie angle in a segment less tlum a semicircle
is greater than a right angle*
Let ABC be a circle, of which BC is a diameter, and E
the centre; and draw CA, dividing the circle into the seg-
ments ABC, ADC, and join BA, AD, DC.
The angle in the semicircle BAC shall be a right angle;
The angle in the segment ABC, which is greater than a
semicircle, shall be less than a right angle;
The angle in the segment ADC, which is less than a semi-
circle, shsJl be greater than a right angle.
Construction. — Join AE, and produce BA to F.
Pboop. — Because EA is equal toEB (I. Def. 15),
The angle EAB is equal to the angle
EBA(L 5);
And, because EA is equal to EC,
The angle EAC is equal to the angle
EC A;
Therefore the whole angle BAC is equal // \ \^ . ^ |^^ +
to the two angles ABC, ACB (Ax. 2). ^ b 7^ ol- '
But F AC, the exterior angle of the tri- \ / i^BcT
angle ABC, is equal to the two angles \. y^ / acb =
ABC, ACB (L 32). ^ an/;^^a
Therefore the angle BAC is equal to the angle FAG rightanete^
(Ax. 1),
284 GEOMETBt.
And therefore each of them is a right angle (I. Def. 10);
Therefore the angle in a semicircle BAG is a right angle.
And because the two angles ABC, BAG, of the triaiigle
ABG, are together less than two right angles (L 17), and
that BAG has been shown to be a right angle;
.*. i. ABC Therefore the angle ABG is less than a right angle,
angi^*^**' Therefore the angle in a segment ABG, greater than a
semicircle, is less than a right angle.
And, because ABGD is a quadrilateral figure in a circle,
any two of its opposite angles are together equal to two right
angles (III 22);
Therefore the angles ABG, ADG are together equal to two
right angles.
But the angle ABG has been shown to be less than a right
angle;
Hence Therefore the angle ADG is greater than a right angle;
^ AP^ ^ Therefore the angle in a segment ADG, less than a semi-
an^e^y circle, is greater than a right angle.
Prop. 82. Therefore, the angle, &c. Q.E.D.
GoBOLLART. — From this demonstration it is manifest that,
if one angle of a txiangle be equal to the other two, it is a
right angle.
For the angle adjacent to it is equal to the same two
angles (I. 32).
And, when the adjacent angles are equal, they are li^
angles (I. def. 10).
Proposition 32. — ^Theorem.
ThA cmgles contained hy a tangent to a circle and a chord
drawn from, the point ofcontaAst a/re equal to the angles in ths
alternate segments of the circle.
Let EE be a tangent to the circle ABGD, and BD a chord
drawn from the point of contact B, cutting the circle.
The angles which BD makes with the tangent EF shall
be equal to the angles in the alternate segments of the aide;
That is, the angle DBF shall be equal to the angle in the
segment BAD, and the angle DBE shall be equal to the
angle in the Begsnen^lidi.
FBOPOSITIONS. 285
CoNSTEUCfPlON. — From the point B draw BA at right
angles to EF (I. 11).
Take any point in the circumference
BD, and join AD, DC, CB.
Proof. — ^Because the straight line EF
touches the circle ABCD at the point B
(Hyp.), and BA is drawn at right angles
'co the tangent from the point of contact
B (Const), __^_-.^-^
The centre of the circle is in BA (III. E ^ b ^ F Thocentn
1Q\ iainBA.
Therefore the angle ADB, being in a semicircle, is a right .*. adb u
angle (IIL 31). . ijlg','
Therefore the other two angles BAD, ABD are equal to a »»*, ._
right angle (I. 32). "i \^t
But ABF is also a right angle (Const) ; \^^^
Therefore the angle A^F is equal to the angles BAD, ABD. = abf.
From each of theseequalstake away the common angle ABD ;
Therefore the remaining angle DBF is equal to the re- .j ^ cad
maining angle BAD, which is in the alternate segment of ~ '^
the circle (Ax. 3).
And because ABCD is a quadrilateral figure in a circle,
the opposite angles BAD, BCD are together equal to two aiso
right angles (III. 22). z BAoi
But tiie angles DBF, DBE are together equal to two 2ri(rht
right angles (L 13); *°»»««
Therefore the angles DBF, DBE are together equal to
the angles BAD, BCD.
And the ansrle DBF has been shown equal to the an£rle = z dbf
BAD; "^ +.DBE
Therefore the remaining angle DBE is equal to the angle . ^ j^^j
BCD, which is in the alternate segment of the circle (Ax. 3). = / bod
Therefore, the angles, ka, Q,E.D.
Proposition 39.— Problem.
Upon a gwen straight line to desemhe a segmmt of a circle,
containing cm cmgle equal to a given rectilineal a/ngle.
Let AB be the given straight fee, anjJ (? the given recti-,
lineal angle, ':^ j^
286
OEOMETItT,
It is required to describe, on the given straiglit line AB, a
segment of a circle, containing an angle equal to the angleO.
Case; I.-*^Let the angle C be a ri^t
angle,
CoifSTBycTiOF.— Bisect AB in F
(I. 10).
-p g From the centre F, at the distance
FB, describe the semicircle AHB.
Tlien AHB shall be the segment required,
Anglo in a PaooF. — Because AHB is a semicircle, the angle AHB
fa?riS? in it is a right angle, and therefore equial to the angle *C
angle. (III. 31).
Case II. — Let C be not a right angle.
Atpoint A Construction. — At the point A, in the straight line AB,
BAD*=c; wake the angle BAD equal to the angle C (li 23)^
J^atriSit From the point A draw AE at right ai^fes to AD (L 11),
angiM to Bisect AB in F (I. 1 0).
From p. From the point F draw FG at right ttngles to AB (L 11),
AB,'dLw and join GB.
perpendi-
cular,
meet!
AEiu
ng
Then
O is the
centre of a
circle
through A
•adB,
Because AF is equal to BF (Const.), and FG is conmloii
to the two triangles AFG, BtGj
The two sides AF, FG are ^iml to th^ t^o sides BF^ FG,
each to each ;
And the angle AFG is equal to the angle BFG (Const);
Therefore the base AG is equal to the base BG (L 4).
And the circle described from tbe dbntre G, at the dis-
tance GA) ^Hll th^r^^fbi^ Ijftss ihtdtigh ^he tk>int B.
Let thhi eifck b» di»8^bed j ^d tet it be AHB.
The IN^99ieHi dkklihaUeohidkimh mfh ^^jiM ti^Aegken
rectilineal cmgh C
PROPOSITIONS. 287
Fboof. — ^Because from the point A, the extremity of the And ad
Uameter AE, AD is drawn at right angles to AE (Const.); the drcie
Therefore AD touches the circle (III. 16, cor.)
Because AB is drawn from the point of contact A, the
angle DAB is equal to the angle in the alternate segment
AHB (in. 32).
But the angle DAB is equal to the angle C (Const.) ; and /. z
Therefore tne angle in the segment AHB is equal to the *^^So
angle C (Ax. 1). . c.
Ther^ore, on the giren straight line AB, the segment
AHB of a circle has been described, containing an angle
equal to the given angle C. Q,£.F,
Proposition 84.— Problem.
From a given circle to cut off* a segment which shall contain
an angle eqtud to a given rectilineal angle.
Let ABC be the given circlei and D the given rectilineal
angle.
It is required to cut off £k>m the circle ABC a segment
that shall contain an angle equal to the angle D.
OoNSTRUCTioK. — Draw the straight line EP touching the Draw tan
circle ABC in the point B (III. ^ . ^""^ ^^
And at the point B, m the y / \ ^^^^^^^-^^.^X
straight line BF, make the angle ^ / ( \ yV ^e
FBO equal to the angle D (1. 23). \ \ /j ^^^
Then the segment BAO shaU \ \ Xy
contain an angle equal to th^ ^^^\r^
ffivenanghD. " * *
Pboof. — Because the straight line £F touches the circle
ABC, and BC is drawn from the point of contact B (Const);
Therefore the angle FBC is equal to the angle in the alter-
nate segment BAC of the circle (III. 32).
But the angle FBO is equal to the angle D (Const.);
Thei'efore the abgle In the segment BAC is equal to the •*- ^ b^
angle D (Ax. 1). =^^f
Thetefbre, mtik the giv^ c^to ABC^ the segment BAC
hi8 h^m imt oSt odntaininf im, angle equal to the given
angle D. Q.E.F.
288
OEOHETBT.
AE - EC.
BE-ED +
EI"^ =
FB« =
AF« =
AE3 +
EfS.
.*. BE'ED
= A£S =
AEEC,
Proposition 35. — Theorem.
If two straight lines within a circle cui one cmotherj th$
rectangle contained by the segm&nis qf one of them shall be
equal to the rectangle contained by the segvMnts of the other.
Let the two straight lines AC, BD cut one another in the
point E, within the circle ABOD.
The rectangle contained bj AE and EC shall be equal to
the rectangle contained bj BE and ED. ■
Case I. — Let AC, B!D pass each of them
through the centre.
Pboof. — Because E is the centre, EA, EB,
EC, ED are aU equal (L de£ 15);
Therefore the rectangle AE, EC is equal
to the rectangle BE, ED.
Case II. — Let one of them, BD, pass through the centre,
and cut the other, AC, which does not pass through the
centre, at right angles, in the point R
CoNSTBUCTiON. — Biscct BD in F, then
E is the centre of the circle; join AJF.
Proof. — Because BD, which pasBea
through the centre, cuts AC, which does
not pass through the centre, at right anglea
in E (Hyp.);
Therefore AE is equal to EC (QL 3).
And because BD is cut into two equal
parts in the point F, and into two un-
equal parts in the point E,
The rectangle BE, ED, together with the square on EF,
is equal to the square on FB (II. 5); that is, the square on
AF.
But the square on AF is equal to the squares on AE, £F
(I. 47);
Therefore the rectangle BE, ED, together with the square
on EF, is equal to the squares on AE, EF (Ax. 1).
Take away the common square on EF;
Then the remaining rectangle BE, ED is equal to the
remaining square on AE; that iS; to tl^ rectfuigle A^^ W)f
mco AE is ecjual V) IBO,
PBOPOSITIONS.
289
' Case HI. — :Let BD, which passes through the centre, cut
the other AC, which does not pass through
the centre, in the point E, but not at
right angles.
Construction. — ^Bisect BD in F, then
P is the centre of the circle.
Join AF, and from F draw FG perpen-
dicular to AC (I. 12).
Proof. — ^Then AC is equal to GC
(in. 3).
Therefore the rectangle AE, EC, together with the square
on EG, is equal to the square on AG (II. 6).
To each of these equals add the square on GF;
Then the rectangle AE, EC, together with the squares on
EG, GF, is equal to the squares on AG, GF (Ax. 2).
But the squares on EG, GF are equal to the square on EF;
And the squares on AG, GF are equal to the square on
AF (I. 47).
Therefore the rectangle AE, EC, together with the square
on EF, is equal to the square on AF; that is, the square on FB.
But the square on FB is equal to the rectangle BE, ED,
together with the square on EF (II. 5) ;
Therefore the rectangle AE, EC, together with the square
on EF, is equal to the rectangle BE, ED, together with the
square on EF.
Take awaj the common square on EF;
And the remaining rectangle AE, EC is equal to the
remaining rectangle BE, ED (Ax. 3).
Case IV. — ^Let neither of the straight
lines AC, BD pass through the centre.
Construction. — ^Take the centre F
(IIL 1), and through E, the intersection
of the lines AC, BD, draw the diameter AN
GEFH.
Proof. — Because the rectangle GE,
EH is equal, as has been shown, to the
rectangle AE, EC, and also to the rectangle BE, ED;
Th^efore ilie rectangle AE, EC is equal to the rectangle
BE, ED (Ax. 1).
Therefore, if two straight lines, &c, Q.EMy
5 rr
BiaeotBD
in F the
centre.
DrawFQ
ftt right
angles to
AC.
.•.AG =
OC.
Now,
AEEC +
= AGa.
.*. AK-EO
+EF« =
AF« = FB«
= BE-EO
+ EF«.
.'. AE'EC
= BE'ED.
Again,
GE-EH =
AEEC and
= DEED,
as Just
shown.
• *• AE*£C
= BE'ED.
21KI
QtOXKTBX.
ADDC +
ECSs
EDS.
.'.AD'DO
+ EB> =
BDS+EB*.
.•. ADDC
= BD«.
DrawEF
perpendi-
cular to
AC.
Propositioii S&^lhecmB.
If/rcmapoijUwUkontachdeiwoaimgkilinegle^m^
one oftDkM euis the eirde, and the oiker Unukee ii/ Ike fwl*
angle eoniained hy the whole line te^aci eute ike drdej and fk
part of it without the circle, diaU he equal to ike equare on tkt
line which touches it
Let J) be any point withoat the circle ABC, and let DGA,
DB be two straight lines drawn from it^ of which DGA cnti
the circle, and DB touches it.
The rectangle AD, DC shall be equal to the square on DE
Case L — ^Let DCA pass through tiie
centre E, and join EB.
Pboof. — ^Then EBD is a lu^t ande
(IIL 18).
And because the straight line AC ii
bisected in E, and produced to D, the rect-
angle AD, DC, together with the square on
EC, is equal to the square on ED (XL 6).
But EC is equal to EB;
Therefore the rectangle AD, DC, togetha
with the square on EB, is equal to llie square
on ED.
But the square on ED is equal to the squares on EB, BD,
because EBD is a right angle (L 47);
Therefore the rectangle AD, DC, toge^et
with the square on EB, is equal to the
squares on EB, BD.
Take awaj the common square on EB;
Then the remaining rectangle AD, DC
is equal to the square on DB (Ax. 3).
Case II. — Let DCA not pass through
the centre of the circle ABC.
CoNSTEUcnoN. — ^Take the centre E
(III. 1), and draw EF perpendicular to
AC (L 12), and join EB, EC, ED.
Proof. — Because the straight line £F,
which passes through the centre, cuts the straight line AC,
which does not pass throTigh the centre, at light angles, it
fdao bisects it (IH. S^ >
PEOPOSITIONS. 291
Therefore AF is equal to FC. ^ ^ fc^^
And because the straight line AC is bisected in F and *
produced to D, the rectangle AD, DC, together with the -^ pcsEp
square on FC, is equal to the square on FD (II. 6). fds.
To each of these equals add the square on FE ;
Therefore the rectangle AD, DC, together with the squares
on CF, FE, is equal to the squares on DF, FE (Ax. 2).
But the squares on CF, FE are equal to the square on CE,
because CFE is a right angle (I. 47) ;
And the squares on DF, FE are equal to the square on
Therefore the rectangle AD, DC, together with the square •^- eS^
on OE, is equal to the square on DE. des.
But CE is equal to BE;
Therefore the rectangle AD, DC, together with the square
on BE, is equal to the square on DE.
But the square on DE is equal to the squares on DB, BE,
because EBD is a right angle (I. 47) ;
Therefore the rectangle AD, DC, together with the square .-. AD-Db
on BE, is equal to the squares on DB, BE. db2+bSi
, Take away the common square on BE ;
Then the remaining rectangle AD, DC is equal to the .j. addc
square on DB (Ax. 3). " ^®'"
Therefore, if from any point, &c. Q, K I).
CoBOLLART. — If from any point without
a circle there be drawn two straight lines
cutting it, as AB, AC, the rectangles con-
tained by the whole lines and the parts of
them without the circle, are equal to one
another; namely, the rectangle BA, AE
is equal to the rectangle CA, AF; for each
of them is equal to the square on the
straight line AJD, which touches the circle.
Proposition 37. — Theorem.
If from a paint vntliout a circle there he drawn tivo straight
lines, one of which cuts the circle j ami tlie other meets it, amd if
the rectangle contained hy the whole line which GuSt& tA^e ca.Tct&>
amd ihepaai of U tvUhout the circle, be equal to tlve squo/r^ cya
292
GEOBIETBY.
Draw DE
touching
the circle.
Then
DE=:DB.
And tri-
angles
DBF and
DEFare
equal in
every re-
spect.
.-.DBF is
aright
angle;
and there-
fore DB
touches
the cir7le.
tlie line which meets tlie circle, the line which meets the circk
shall touch it.
Let any point D be taken without the circle ABC, and
from it let two straight lines, DCA, DB, be drawn, of which
DCA cuts the circle, and DB meets it; and let the rectangle
AD, DC be equal to the square on DB.
Then DB shall touch the circle.
Construction. — ^Draw the straight line DE, touching the
circle ABC (tH. 17);
Find F the centre (III. 1^ and join FB,
FD, FE.
Proof. — ^Then FED is a right angle
(III 18).
And because DE touches the circle ABC,
and DCA cuts it, the rectangle AD, DC
is equal to the square on DE (III. 36).
But the rectangle AD, DC is equal to
the square on DB (Hyp.);
Therefore the square on DE is equal to
the square on DB (Ax. 1);
Therefore the straight line DE is equal to tJie straight line
DB.
And EF is equal to BF (I. Def 15);
Therefore the two sides DE, EF are equal to the two
sides DB, BF, each to each ;
And the base DF is common to the two triangles DEF,
DBF;
Therefore the angle DEF is equal to the angle DBF (L 8).
But DEF is a right angle (Const.) ;
Therefore also DBF is a right angle (Ax. 1).
And BF, if produced, is a diameter; and the straight line
which is drawn at right angles to a diameter, from the ex-
tremity of it, touches the circle (HE. 16, Cor.);
Therefore DB touches the circle ABC.
Therefore, if from a point, &c. Q,E,D.
fiX£BClS£S OK THE PBOPOSITIOKS. 293
EXERCISES ON BOOK III.
Prop. 1—15.
1. Two straight lines intersect. Describe a circle passing thronsli
the point of intersection and two other points, one in each straight
line.
2. If two circles cut each other, any two parallel straight lines
drawn through the points of section to cut the circumferences are
equaL
3. Show that the centre of a circle may be found by drawing per-
pendiculars from the middle points of any two chords.
^ 4. Tlurough a given point, which is not the centre, draw the least
line to meet the circumference of a given circle, whether the given
point be within or without the circle.
5. The sum of the squares of any two chords in a circle, together
with four times the sum of the squares of the perpendiculars on them
from the centre, is equal to twice the square of the diameter.
6. With a given radius, describe a circle passing through the
centre of a given circle and a point in its circumiierenee.
7. If two chords of a circle are given in magnitude and position,
describe the circle.
, 8. Describe a circle which shall touch a given circle in a given
point, and shall also touch another given circle.
9. If, from any point in the diameter of a circle, straight lines be
drawn to the extremities of a parallel chord, the squares of these
lines are together equid to the squares of the segments into which
the diameter is divided.
10. If two circles touch each other externally, and parallel dia-
meters be drawn, the straight line joining extremities of these dia-
meters will pass through the point of contact.
11. Draw three circles of given radii touching each other.
12. If a circle of constat radius touch a given circle, it will
always touch the same concentric circle.
13. If a chord of constant length be inscribed in a circle, it will
always touch the same concentric circle.
14. The locus of the middle points of chords parallel to a given
straight line is a line drawn through the centre perpendicular to the
parallel chords.
Prop. 16—30.
15. Show that the two tangents from an external point are equal
in length.
16. Draw a'tangent to a given circle, making a given angle with a
given straight line.
17. If a polygon having an even numoer of Bide& \)^ \XAQt{^^ Vdl ^
circle, the ^912109 of the alternate angles are equs^*
294 GEOMETBT.
18. If sncli a polygon be described aboat a circle, the snins of the
alternate sides are each eqnai to half the perimeter of the polyi^n.
19. If a polygon be inaoibed in a circle^ the sum of the angles in
the s^ments exterior to the polygon, together with two ru;ht
andies, ia equal to twice as many right angles as the pcdygon has
20. Draw the common tangents to two civen circles.
21. From a given point draw a straight line cutting a ^tven drdei
so that the intercepted s^ment of the line may have a given length.
22. The strai^t line which joins the extremities of equal arcs
towards the same parts are paralleL
23. Anv paralleiogram described about a circle is equilateral, and
any parallelogram inscribed in a circle is rectangular.
24. Two opposite sides of a quadrilateral circumacribinc a dicle
touch the circle at extremities of a diameter. Show that t£e area of
the quadrilateral is equal to one-half the rectangle contained by the
diameter, and the sum of the other sides.
Pkop. 31—37.
25. A tangent is drawn to a circle of 21 inches diameter from i
point 17*5 inches from the centre. Find the length of the tangent
26. Show that a man 6 feet hi^ standing at the sea level, has a
view of 3 miles (approximately) m every direction, along a horizontal
plane passing through his eye.
27. The angle between a tangent to a circle and the chord thlxragli
the point of contact is equal to half the angle whidi the chord sob*
tends at the centre.
28. From a given point P, within or without a circle, draw a
straight line cutting the circle in A and B such that PA shall be
three-fourths of PB.
Ex. Let the circle be of 1*5 inches radius, and point P 3*5 inches
from its centre. Prove your construction by scale.
29. The greatest rectaSogle which can be inscribed in a circle is a
square whose area is equal to half that of the square described upon
the diameter as side.
30. If the base and vertical angle of a triangle remain constant in
magnitude while the sides vary, show that the locus of the middle
point of the base is a circle.
31. Given the vertical angle, the difference of the two sides con*
taining it, and the difference of -the segments of the base made by a
perpendicular from the vertex, to construct the triangle.
32. Show that the locus of the middle point of a straight line,
which moves with its extremities upon two straight lines at right
angles to each other, is a circle.
33. Show how to produce a straight line, that the rectangle con*
tained by the given line, and the whole Une thus produced, may be
equal to the square of the part produced.
Ex. If the leng^li oit\i<& ^NQuline be 2 inches, show geomeiricallif
that the lei^li oi tiiQ i^%c^ ^x^u<;i^Ss^ VnI^ -v \^\ssis^qi«u
EXERCISES ON tHil ]PROP6smO^S. ^9^
34. Given the height and chord of a segment of a circle to find the
I'adins of the circle.
Ex. If the chord be 24 inches, and the height of the segment be 4
inches, show that the radius of tiiie circle is 20 inches.
35. Show that the locus of the middle points of chords which pass
throogh a fixed point is the circle described as diameter upon the
line joining the fixed point and the centre of the fpyea circle.
36. Let AGDB be a semicircle whose diameter is AB, and AD, BO
any two chords intersecting in P; prove that
AB« = DA-AP + CB-BP.
MATHEMATICS.
SECOND STAGE.
SECTION 'II.
ALGEBRA.
CHAPTER I.
QUADRATIC EQUATIONS.
1. Equations of this class, when reduced to a rational in-
tegral form, contain the squa/re of the unknown quantity, but
no higher powers.
When the equation contains the square ordy of the unknown
quantity, and not the first powsTy it is called a pwre quadraJIk,
Thus, «* - 25 = 0, 4ar* + 10 = 19, 5a? = 180, are/?u«
quadratics. When the equation contains the squ>are of the
unknown quantity, oa weU as the fi/rst power, it is called an
ad/ected qtuidraUc.
Thus, sc" - 5a; = 6,a2-a;-30=0, 2ar» + aj + 3 = 6,
are adfected quadratics.
Pure Quadratics.
2. To solve these, we treat them eicactly as we do simple
equations, until we obtain the value of the square of the un-
known quantity *, then, taking the square root of each side,
we obtain the 'oal'm oi ^<& xccSBCkSswcL o^^^so^i^. It will be
ADFECTED QUADRATICS. 297
seen (Stage I., Alg. ; Art. 35) that the unknown quantity in a
pure quadititic has always two values, egrual in magnitude^
but of opposite sign.
Ex. 1. Given 3 a^ + 12 = 687, find x.
We have 3a^ = 687 - 12 = 675, or a;* = 225.
.\ X = ± 15.
^ « ^. 2aj + 7 2aj - 7 66 - ,
Ex. 2. Given ^^-^ - ^^^^^ = ^-^^99, findo:.
Bringing the fractions on the first side to a common de«
nominator, we have —
(2a; + 7)^ - {2x - 7 )^ _ 56
jc(4ic2 - 49) ~ 6aP - 99'
56 X 56
.^^' x{^^ - 49) "^ 6aj» - 99'
^^' 4ar' - 49 "^ 6 ar^ - 99 ^ clearing effractions;
then, 6aj^-99 = 4a?-49, from which
aj2 = 25
a; = ± 5.
Adfected Quadratics.
3. Solution by compkting the sqtia/re.
Suppose we have given the equation a^ + 2 ooj = 3 a^, to
find a;.
It will be remembered that {ixP + 2 ax + a?) is a perfect
sqiuj/re, viz., the square of {x + a). It is evident, then, that the
first side of our equation will become a perfect square by the
addition of a^ as a third term.
Adding, then, c^ to each side of the given equation, we
have—
a? ■{■ 2 oa? + a* = 4 a^, or
(x + a)^ = 4 a*.
Taking now the square root of each side, we have-«
35 + a = ± 2 a.
,\ X = ± 2 a — a.
s= a or - 3 tt.
298 ALGEBRA.
We may remark that the quantity a*, added to the expres-
sion as" + 2 005 in order to make it a pefirfect aqua/re, is the
square of half the coefficient of x* llie operation itself is
called completing the aquxire.
An adfected qua(&atic may therefore be solved as
follows: —
1. Bedtice arid arrange U until aU the terms involving x
are on the first side, and the coefficient qfi^is unity,
2. Complete the square by adding to each side the
SQUARE OF HALF the coefficient qfx»
3. Take the squa/re root of eaxih side, piU a dovhle Sign to
the second side, a/nd transpose the term of the first side not
irvDolving x.
Ex. 1. Solve the equation 3 a^ + ISoj + 4 = 52.
We have Za? + 18aj = 52 - 4= 48; or, dividing each
side by 3, as" + 6 a =16.
Here, the coefficient of a; is 6, the half of which is 3<
Adding then the square of 3 to each idde, to complete thi
square, we have—
aj" + 6aj + 3" = 16 + 9 = 25.
Taking the square root ot each side, w€l have —
aj + 3 = ±5;
.-. a: = ± 5 - 3 = 2 or - 8.
•ci^on* a; + 3 a;+l 4aJ + 9 l2aj + Vt
Ex. 2; Given j -^ = ^r = - -^ r^^
aj + 4: aj+2 2aj + 7 6a; + 16
find X.
WehaveA - -JL_\ A ^ ^ \ = A - o— ^1
V aj + 4^ V a; + 2/ \ 2a; + 7/
" r " 6 a: + le)^
br, simplify^/^ " ^ = 6^6 " jAi'
, ; (x-h 4) - ta; + 2) ^ 15(2a;+ 7) - 5(6a;+ 16) .
^^\ (x + 2) (a; + 4) (6aj + 16) (2aj + 7) '
^ — — _ - ] 2 25
ADFECTED QlTADRATICS. 299
or, clearing of fractions —
24ar^ + U8aj + 224 = 25ar» + 150a; + 200;
or, transposing and reducing, -cc^-2a3= — 24;
or, changing the sign of each side, then —
aj* + 2aj = 24;
or, completing the square, a:? + 2 05 + 1* = 24 + 1 = 25.
Taking the square root of each side, we have —
jc + 1 = ±5
4aj=±5-l=4or-6. i
Ex. 3. Solve the equation a? + 6 a; + 25 = 0.
We haveaj* + 6 a? = - 25;
or, completing the square — . ,..
a? + 6a; + 32 = - 25 + 9 = - 16;
or, extracting the square root of each side —
a; + 3 = ± Jll^
:,x= - 3 ± J~rTQ,
As the quantity ^/ - 16 has nd exact or approximate
value, the given equation has no real rootSi The roots are
therefore said to be imaginary.
4, SoitUion hf htealcmg mtofactors*
We have seen (Stage I., Alg., Art* 30) that it Id often
easy to find by insped;ion the fieictors of quadratic express
sions. We may make use of this knowledge to solVe quad-
ratio equations.
Ex. 1. Solve the equatioh a;^ -f Sa;== 66.
Transposing all to lie first side, we have —
ai» + 5 a; - 66 = 0;
And, resolving the first side into its elementary factors, we
get—
(a; - 6) (a; + 11) = 0.
Now, if either of these factors is put equal to 0, the
equation is satisfied.
Hence, we have, a; - 6 = 0, and a; + 11 = 0;
or, a; = 6, and a; = - 11.
.*. 6 and - 11 are the roots required.
300
Ex. 2, Oiren a^ - (a -¥ h)x -i- ab = Oj find ae.
We hare (« - a) (« - 6) = 0.
Now tiiis equatioii is natiiififd bj malring cidftff of tiie
factors s 0.
Hence, fl5-o = 0, ora5 = a;)
and, a;--6 = 0, ora5 = 6;j
/. a, i are the roots required.
Ex. L
1,
4a? - 7 = 29.
2. 5a? + 6 = 86.
3.
3«» - 7 oa'
—2 = ^*'
4, |a? + f = 6.
5.
3a! - 1^ - 0.
X
6. 2 (a» + i») - a
7. ♦ajN/24 + ar^ = 4 + aj».
8. * \/a - a = Va - a/^* + a^. *
10. 1 ^ 1 -^ ^
,0 ® + Vaj"- 1
^"^^ /-a i- «= «• 13. ar* - 5a; + 6 = 0.
U. ar" - a = 72. 15. 3ar^ - aj = 2.
10. 4a? - Oa = 28. 17. Gar* + a; - 35 = 0.
18. a:* + 6ix + 8 = 0. Id. aa^ + hx + c = 0.
20. 3«- H-^ =4a;-7. 21. a - ^ = 1.
2 a? - 3. X
22, 2»+ 1 - ^^ = 11.
2a; + 1
ADPECTED QUADRATICS. * 301
23. (a + bx) (ex + d) == (a + b) (c + dx).
24. (a - 6) (ar^ - 6^) = (a + 6) (aj - 6)2
25. (a + 6) (05 — a) (x - h) - ahx.
Q/. a ^ a; 6 , a; , a^ - 6^
Jb. — + -.=_ + _ + ^ — ,
a; 6 a; a ab
21. aa? - (c + (I)x ^ht? -^.
a — b
28. aV - (a - 6)2 aj + a» - 6W + (a^ + a6 + ft^js^j + b\
29. Ji-.-^ = 3i.
a + 2 X ^
30 ^ "^ ^ + -L = ^^ + ^
* 2aj + 3 13 4aj-5*
31. 2^_±_i + jp-^— - = 61.
aj 2aj + 9 ^
32. ?-i4 + ^-iJ^ = 0.
a; + 3 a; + 8
„o 3a;-7^5aj+3 5
4aj - 2 7a; + 4 9
34 ^^ + ^ . 4a; - 1 _ 7 a; + 1
* 2 a; - 7 x - 2 "" a; - 3 *
35 ^^ + ^ _ a; + 2 _ 7 a; + 8
a; + 2'"4a; + 4~ 4a; + 13*
36 ^^ - 2a; + 9 ^ 2a; + 24
'a;-2~ a;+2 " 2a; -1"
o^ 3a;2 _ 2a; + 7 ^ ar^ + 7a; - 3
6a^ - 4a; + 11 2ar» + 14a; - 9'
,Q3a^+10 ar» + 2a; + 3 2ar» + 2a;+10
j». — _ — = ,
X X + 2 a; + 1
39. * ^/5a; + 6 + Jx + 10 = 4.
40. >/3a; - 4 - j2x~^n^='Jx^
41. * J^fTWV^ + Va^ + 62 - a; = ^/?i!?.
-■ N 6 •
* See I(emark, pa^^l^
302 ALGEBBA,
42. * V2 a + a? + V4 a - a? = 2 ijx - a.
43. * si ax + 6 + ijbx + a = v (a - h)x + a + 6 + 2 Voi^
46. (a - 6) (ic + 6 - c) (a5 + c - a) + a^(a3 + 6 — c) +
b%x + c " a) + <?{x + a - 6 = 0).
47. Show that, if a; be real, the value of a; + - cannot lie
X
between 2 and - 2.
a?
48. If, in the equation a^ - OQ ~ ^' *^® quantity a; be real,
Bhow that (I cannot be greater than 5.
Equations which may be Solved like Quadratics.
6. Certain equations of a degree higher than the second
may be solved like quadratics. It will be seen that, although
it is impossible to lay down definite rules for the treatment
of every such equation, the object to be attained is either
(1.) To throw the equation into the form —
when X is an expression containing the unknown quantity,
and solving when possible this equation for X; or (2.), To
strike out from each side a factor containing the unknovn
quantity, thus reducing the dimensions of the equation,
and obtaining a value or values of the unknown quantity
by equating this factor to zero.
Ex. 1. Solve the equation aj* + 144 = 25 a;^.
We have, transposing, aJ*-25a^= -144;
or, (iB«)2 - 25 (a^) = - 144;
JSQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 303
or, completing the square —
{a?f - 25 (a:*) + (V)' = ^ - 144 = V-
.-. ar» - jy» = ± J.
/. ar^ = V ± 1 = 16 or 9.
Hence, x ~ ± 4 or ± 3.
The given equation has therefore four roots, being as many-
roots as the degree of the equation.
Ex. 2. Solve a/* + 3 a^ = 88.
We have, {aFf + 3 (ar*) = 88 ; or, transposing,
{oi?f + 3(aj») - 88 = 0;
or, (a^ - 8) (a:» + 11) = 0.
Hence, the given equation is satisfied by —
a? — 8 = 0, and also by a^ + 11 = 0.
We have then a' - 2» = 0, and ar* + ( ^ll)' = ;
or, breaking into factors, we have —
(aj - 2) (ar» + a • 2 + 2*) = 0, and
(« + ^|a? - a;-^ir+ (^11)'} =0.
From the^^i of these equations, we get —
aj-2 = 0ora: = 2,
and a* + 2a; + 43:0, from which two other roots may be
found.
And from the second equation, we get —
aj + ^11 = 0, ora = - «/ll,
and a? - X ^/W -{• sl\2\ = 0, which gives two other roots.
We have therefore shown how to obtain the six roots of
the given equation.
Ex. 3. Solve —
23a» - 75a; - 6ajA/4a« - 9a; + 9 + 40 = 0.
Changing the sign of each side, and transposing, we have —
exj4^ - 9a;+ 9 = 23a? --Idx + 40;
adding to each side 13a^ - 9a; + 9, then —
13a? - 9a; + 9 + 6xjia? - 9a; + 9 = 36ai'- 84a; +49;
or, {ia? - 9 a; + 9) + 2 (3a;) >Jia? - 9a; + 9 + (3a;)'
= {Qxy - 2(6a;)-7 + V\
304 ALGEBRA.
or, (a/4jb^ - 9x + d + Sxf = (6a; - 7)1
.-. V4ic2_ 9a. + 9 + Sx = ± (6a-7), (1.);
or, taking the upper sign —
A/4ar» - 9a; + 9 = (6a; - 7) - 3a; = 3 - 7.
Hence, squaring eacli side—
4ai" - 9a; + 9 = 9ar^ - 42a; + 49;
or, 5 ai* - 33 a; + 40 = ;
or, {x - 5) (5 a; - 8) = 0.
/. a; = 5 or f .
Again, taking the lower sign of ( 1), we have—
V4ar^-9a; + 9= -(6a; - 7) - 3aj = - 9a; + 7 (2.);
or, squaring —
4a;» - 9a; + 9 = 81a? - 126a; + 49;
from which 77 ar» - 117a; + 40 = 0;
or, (a; - l)(77a; - 40) = 0.
.-. a; = 1 or 4^.
Hence, the roots of the given equation are 5, |, 1, ^,
Kemare. — ^If we proceed to verify these values of a;, we
shall find that the last two values — ^viz., 1 and 4? — ^will not
satisfy the given equation unless we obtain the value of
iji a;^ - 9 a; + 9 by means of the equation from which these
last roots were found.
Thus from ( 2 ) we find, on putting 1 and 4^ successively
for X —
V4ar» - 9a; + 9 = - 9 (1) + 7 = - 2,
and V4ar» - 9a; + 9 = - 9 {^) + 7 = 2^;
and on substituting either of these values of \/4a;* - 9a; + 9
along with the corresponding value of x, the given equation
is satisfied.
Ex. 4. Solve —
(a; + 6 + c) (a; + 6 — c) (6 + c - a;)
= (a + 5 + c) (a + 6 - c) (6 + c - a).
By inspection we see that a is one of the roots. We shall
therefore so arrange the equation as to be able to strike out
^ — a as a factoi of eadi «ide (Art 5),
EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 305
We have —
(x + b + c) (x + b " e) _ 6 + c - a.
(a + b + c){a + b - c) b + c - x'
(x + by - (^ _ b + c " a
^' {a + by - c* " b + C--X'
or, taking the diffireme of numerator and denominator —
(x + by - (a + by _^ X - a_ ^
or,
{a + by - c^ b + c — x'
(a; - a) (aj + 2 6 — a) _ x - a
{a + by " <^ b + c - X
Dividing each side by a; - a, we have —
X + 2b " a _ 1
(^hy^c'" b +C--X' and a - a = 0, oraj = a;
Hence also (aj + 2 5 - a) (5 + c — a;) = (a + 5)' — c^; an
ordinary quadratic from which two other roots may be
determined.
Ex, 5. * Solve the equation —
a'6V - ^a^ «*« = (a - 6)V.
i_
Dividing each side by a?^, we have —
€?Vx^^ - 4 a*6*a: 2^'"= {a - by ;
or, (a&»*w^ - 4 a*6^ (abx^) = (a - «>)- ;
or, completing the square —
\absf^) - 4 a^ft* {ahx^ + (2 aij*) ;
= (a - 6)2 + 4 06 = (a + 6)'.
* We shall luwmt in thia example that the lawa of multiplication
«nd division proved in Algebra, Stage I., Arts. 26, 87i hold for feaa-
tdonal indices,
9 V
<»:.
x^
9
^^^ ALGEBRA,
■''■''■
Hence, extracting the square root—
.% a6«^ = ± <a + 5) + 2a*6i = (ai + 6*)«or-(ai-6^)'i
Then, raising each side to the i^pq)^ power, ve have—
(1 \\^P^ /I l\*i>«
Henoe, taking the (^ - ^)th root, we get —
/I 1 \±PSL /I 1 \J£
/I i\i^
We have also, since the &ctor a^ has been struck out-*
x^ = 0, and .-. a? = as another solution,
Ex. 11.
1. 4 aj* - 11 ar» = 226.
2. 5 »» - 17 ar» = 184.
3. aj» + 5 »-« « 25^.
4. (a? + S» + 3)« + 2ai* = 189 - 6«.
b. 3aj-*^-.*17«-« = 1450.
'':lfc y£TT 2 + vo? + 12 = 6,
\.^ + 9* + 20 = aj* + 9.
EQUATIONS WHICH MAY BK SOLVED LIKE QUADRATICS. 307
0. (3 a? + 4) + N/3a + 4 - 12 = 0.
2
1. or* .+ 2 a5 = — 7-T - 1.
aj + 4
o ^/a - X si a - cc \^iK
z, . ■■ •- = . — •
X a a
- 4 1
3. 9 aj + 24 J» - -7= /'-i^ + 9 ) = 65.
4. a; = >/a * fjb - x -^ *JS • Va; + a.
1 1 ^ a^ + 2^
6 ~ - 2 /^^ - ax + a? _ ^^ _ i
Q fa* + a? / a? ,
9. («♦" + 1) («* - 1)» = 2 (oj + 1).
a^ _ 975 ^_-_9 7 _ ()
-"•^^Tg 256-~l?^ ^g'^-
aj+lla + 1 ajj ar
22. 6ar^ - 5aj-8V3a» + 6aj - 4 « 12.
23. 2aj» + 6a; - 2xj2a^ - 7aj + 1 =» 35.
24. 20aj* - 9aa?-8a;V5ar»-3aaj + 2a^ = 7 a\
25. ar^(» - 2)* + ea^{x - 2) =t 24» + 36 - So'.
26. 2ar^ + 20aj - V3a:> - a: + 5 ^ 105.
27. 4a? + 12aj - 2xJ^9^ - 2a> + 19 = 30,
28. «^ + o^ + 1 = a(»' + « + 1).
29. a^ - 1 = 0.
• f
308 ALGEBRA.
30. (ar» - af + (ar» - hf + (ar^ - c)»
= 3 (ar* - a) (a:» - 6) (or* - c) + 9 oaj* - a(a + 6 + cf.
31. (a? - b) (x^ - c^) = (<i - 5) (a'^ - c»).
32. (as - a) (aj — 6) (a; — c) = (m — a) (m — 5) (m - c).
33. (aj - a) (aj - 6) (aj — c) = (a + 6) (a + c) (6 + c).
34. (n - l)a; (a^ + oa; + o*) - a^ - a?,
35. (aaj - 6)» + (caj - cZ)^ = (a + c)|a:» - (6 + (f)».
36. ar»" +ar»".l'+ 1 = a(aj8 + a; + 1).
Simultaneous Quadratic Equations.
6. The following worked examples are given as speci-
mens of the methods to be employed, but it must be under-
stood that^ practice alone will give the student complete
mastery over equations of this class.
Ex. 1. Solvear^ + j/" = 20 (1.) 1
X +y = 6 (2.)/
As we have given the 8vm of the unknown quantities^ we
shall work for the difference.
From (2), multiplying each side by 2, we have—
2a? + 2y* ^ 40
and from ( 1 ), squaring, a? + 2xy + y* = 36
Then, subtracting, a?-2ajy+ y* = 4;
and, taking the square root, we have oj - y = + 2 (3).
(2) + (3^, then 2a; = 8 or 4, and .-. a; = 4 or 2.
(2) - (3), then 2y = 4 or 8, and .*. y = 2 or 4.
Note. — ^Having found that a; = 4, ^ = 2, we might have told by
inspection that the values x = 2, y = 4, would also satisfy the giyeu
equations, for x and y are similarly involved in both equations.
Ex. 2. Solve a? - 3/* = 6 (1.) )
«^=6 (2.)/
As we have given here the difference of Ihe squares of the
unknown quantities, it will be couvenieQt to work for tb^
9tm of the «guare8.
SIMULTANEOUS QUADRATIC EQUATIONS. 309
From (1.), squaring, aJ* - ^o^y^ + y* = 25,
and ffom (2.), squaring, &c., 4af^y^ =144
Then, adding, sc* + 2 ar^y^ + 3/* = 169
and taking the square root, a* + j/^ _ + _ 13 ^3^^
(3.) + (1.), then, 2 ar* = 18 or - 8, or ar» = 9 or - 4,
and .•. a; = ± 3 or ± 2 J - I.
(3.) - (1.), then, 2 3/^ = 8 or - 18, or 2/* = 4 or - 9,
and /. y = ± 2 or ± 3 V - 1.
Note. — The Btadent will see that the pairs of valaes which s atisfy
the given equations are, x — S,y =2; x—- 3, y = -2;a; = 2 V ~ 1»
y = 3V-l;aj= -2V-1, y= - SnT^.
Ex. 3. Solve a? + y = 11a? (1.)) .
3/^ + 0?= 11 y (2.)/
Subtracting, then, as'-y* — aj + y = llaj- 11 y;
or, a;* - ^ = 12 (a? - y).
Now (a; — y) is a factor of each side, and hence, striking it
out, we have —
aj + y = 12 m,
and also a; - y = (4.).
Equations (3.) and (4.) may not be used as simultaneous
equations, but each qf them may be used in turn with either
of the given equations.
Thus, taking equations (3.) and (1.), we have —
(1.) - (3.), ar'-a;= 11a; - 12,
from which a; = 6 ± 2 sf^;
and hence from (3.), by substitution, we easily get-
y = 6 + 2 Ve:
Again, taking equations (4.) and (1.) —
we have, from (4.) x = y,
and .*. from (1.), ar* + a; = 11 a; or a;^ = 10 a?,
from which a; = 10 or ;
and so, from (4.), y = 10 or ^»
310 ALG£filtA.
Hence, the pairs of values satisfying the given equations
are —
X = 10,y = 10;a; = 0, y = 0;
a; = 6 + 2 J^y_y =6-^2 »J^j
a; = 6 - 2 V6, y = 6 + 2 V«-
Note. — It is worth while remarking that when each of the terms
of the mven equations contain at least one of the unknown quantities,
the vames x = 0, y ^ will alwa^ satisfy.
Ex. 4. Solve 3ar» - 2a^ = 55 (1.))
ar» - 5icy + 82^ = 7 (2.)/
Multiplying the equations together crosswiae^ we get —
55ai» - 275ajy + 440 y« = 21 a? - 14a?y;
or, transposing, 34 ar* - 261 xy + 440 y* = j
or, (2a; - 5y)(17aj - 88 y) = 0,
from which 2 a; = 5 y, and 17 a; = 88 y.
Each of these equations taken in turn with either (1.) or
(2) will easily give the required values of x and y.
Ex. 5. aj* + y* = 337 (1.),
a; + y = 7 (2.)
From ( 2 ), raising each side to iih& fourth power, we have—
a^ + 4ar'y + Gar'y^ + 4:xf + y* = 2401; (3.)
(3) - (l),then4a»y + 6afy + 4a^ = 2064;
or, 2ar'y + 3ar'y» + 2 0^ ^ 1032;
or, arranging, 2 xi/{x + yf - ar'y^ = 1032 ;
but from (2), (x ^ yf =» 49,
and hence, 3 ajy(49) - ix?f =^ 1032 ;
or,a^y2 - 98ajy + 1032 = 0,
from which xy = 12 or 86, (4.)
From ( 2) and (4), x •- y may now be easily obtained, and
hence also the required values of x and y.
Ex. IIL
1. a; + y = 5, a^ = 6.
2. a; - y = 2, ajy = 15.
^. at? + if = 2^,xy =^ 13.
4. a? + 2/^ = 20, X ^ ^ = ^
filMULT»ANEOlJS QtrABRATIO EQUATIONS. 311
5. ar^ + 3r» = 29, a: - y = 3.
6. ar» - 2r» = 13, {x - yY = 1.
7. ar» - 3/* = 27, ccy = 18.
8. ar^ - y2 = 12, a; + y = 6.
9. ar* + 2/« = 63, aj» - s^ = - 46.
10. a^ + a^ = 28, 2/^ + ajy = 21. ' "
11. ar* + ay + y2 == 19^ a^ _ ^ _. _ 3^
12. a; + y = 13, Jx + ^y = 6.
13. a* + a^ + y* = 84, a; + ^.^ + y = 14.
14. a^ + y3 = 35, ar'y + ay» = 30.
15. - + -- = a, -_ + ^ = 5.
X y ar y*
16. a? + y = o(a; - y), a* + y* = 6^.
17. a^ - y* = a, ai^ - y2 = 6.
18. a: + y = 6, a* + y* = 36.
19. a; + y = 6, a:' + y» = 275.
20. ar» + y2= V(aJ + y), sey = 6.
21. a: - y = 2, a» - 3^ = 98.
22. a^(aj + y) = <«, ar*y*(ar* + y^^ = 6.
23. a:y(a; + y) = 30, ar'y^(a!» + y*) = 9900.
24. 4ai" - 3ay = 18, 6y» - 2a^ = 8.
25. «* + «* = 35, ai + yi =— r-r
26. a;^ + yJ = 3 x, «4 + y» = x.
27. ay + 6 = ?^, a; + y + 4 - ^^
ay X + y
28. (aj + y)2 + 2(aj - y)^ = 3 (a; + y) (a; - y), ai* + y* = 10.
29. a^ + lOay + y^ = \^ {a^ - y^, a? + 5f = x + 13y.
30. a^ - 2ar'y2 + y* = 1 + 4»y, ar*(a5 + 1) + y^(^ + 1)
= ay.
«, o o 9aj"-y«-117 8ar»-y*+l
31. 3a; + y - 9 = —5 ^ ^— = -. — . ^ ^ ^ .
^ 3a;— y — 6 4x + ^ -V \
312 ALGEBRA.
y-2 a;-2 y + 2 a!-2 *
33. ^^^4±f^) = 3^.2V2,i-^=A.
34:. y ^2 J5x^ + y + 3 = 32 - 5a?,
/^?.+ n* + f5. + if^ = 144.
\y a;/ V x/^
36. (aj + y) ajy = c (&c + ay),
xy {bx + ay) = ar^^ + ahc (a; + y — c).
37. ^ = a?(ay - 5a;), a^ = ax - by.
38. n/^^^TT -f 1 ^JfTg^-^ %(y + 1)^ = 36(2^ 4- V)-
39. 05 = , y - , z =
y + « a; + » X + y
40. a; + y + « = 6, a^y + a» + y« = 11, ajy« - 6.
il. a? - yz = 0, y^ - xz = Oy s^ " xy == 0.
42. ay» = a\x + y) = l^{y + «) = c^(a; + z).
a' 52 ^a
43. «== + j/» + «» = ^ = P = ^.
44. a; + y + « + «4 = 4a + 46,
a^ + a:!2j + a»^ + ^« + ^+«t« = 6a*+ 12 a5 +65',
02/^ + 25^ + xzuu + 2/2^ = 4a^ + 12a% + 12ai* + ih\
ocyzu = a* + 4a®6 + 6a^6^ + 4a6' + 6*.
45. a?y^ + aj^/^^ + aryz = a,
2/*«^ + a^« + a:y«^ = 6,
03?^ + ar^y^ + a:y»* = c.
46. (a; + 2/)« + i2? = 1125,
aj + y + « = 15,
a?y = 24.
Ill
47. If oa;* = W = cs^» and - + - + - = a, show that
•^ a; ^ s;
oar* + 62/* + c»^ = ^ci^ ^ ^^ -v c* )» a\
JPROBLEMS PRODUCINa QUADRATIC EQUATIONS. 313
48. Given j^=l+r, P=-(l- E'% M = PBT,
show that i? = -41 p - t7) + 1.
CHAPTER II.
Problems Producing Quadratic Equations.
7. We shall now discuss one or two problems whose solu-
tions depend upon quadratic equations.
Ex. 1. A person raised his goods a certain rate per cent.,
and found that to bring them back to the original price he
must lower them 3 J less per cent, than he had raised them.
Find the original rise per cent.
Let X = flie original rise per cent.,
X
then fAQ-T — • 100 = the faU per cent, to bring them to
the original price.
Hence, by problem —
aj = 20 or - 16|.
The value a; = 20 ia alone applicable to the problem. Re-
membering, however, the algebraical meaning of the negative
sign, it is easy to see that the second value, x = — 16|,
gives us the solution of the following problem: —
A person lowered his goods a certain rate per cent., and
found that to bring them back to the original price he must
raise them 3^ more per cent, than he had lowered them.
Find the original fall per cent.
The above solution tells us that the fall required is 16f
per cent.
Had we worked the latter problem first, we should have
obtained a; = 16f or - 20, the value a = - 20 indicating
the solution of the former problem.
314 ALGEBRA.
Ex. 2. Find a number such that when multiplied by its
deficiency from 100 the product is 196.
Let X = the number,
then 100 - 05 = its deficiency fi:t)m 100.
Hence, by problem —
X (100 - a) = 196, 6r ar» - 100 a; + 196 = 0;
from which, aj = 2 or 98.
Both these values will be found to be consistent with the
conditions of the problem.
Ex. 3. The number of men required to build a house is such
that, when four times the number is subtracted from three
times the square of the number, the result is 160. Find the
number of men.
Let X = the number of men,
then, by problem —
3 0? - 4 aj = 160, fipom which
a: = 8 or - 6f.
The value as = 8 is alone applicable to the problem as it
stands. If, however, we may coaiceive of a./racti<mai number
of men-and this we may easOy do here by supposing a ^s
work to be equal to { of a man's — we find that the second
result gives us the solution of the following problem : —
The number of men required to build a house is such that
when four times the number is added to three times the
square of the number, the result is 160. Find the number.
The answer, as above indicated, is 6 men and 1 boy,
where a boy is worth } of a man.
The student will find, however, that in some cases there is
no ohmous interpretation to the second result^ owing oocasiim-
ally to the fact that certain term$ are luaed in the problem
to which the results will not aj^ly, and indeed liiat the
algebraical expression of the conditions of the problem is
more general than the language of the problem itself..
Ex. lY.
1. Find a number whose square is equal to the product of
two other nimibers, one of which is less by 6 than the required
iiumber, and the o^qx g;i:e«.\«i:\i^ ^ -^vn twice that numb^.
PROBLEMS tROBUCINO QUADRATIC EQUATIONS. 315
2. When the mimerator and denominator of a certain
fraction are each increased by unity the fraction is increased
by tJ^^, and when they are each diminished by imity the
fraction is diminished. by ^. Find the fraction.
3. The mean proportional between the excess of a certain
number above 21, and its defect from 37, is 28. Find the
number.
4. A number of articles, which were bought for £4, cost
each 3 shillings more than half as many shilHnga as there
were articles. Find the number of articles.
5. There is a square court-yard, such that if its length be
increased by 10 feet, and its breadth diminished by 20 feet,
its area woidd be 5,104 feet. Required the side of the square.
6. If the number of shillings given for an article be added
to the number of articles which can be bought at the same
price for 18 shillings, the result is 11. Find the price.
7. Two travellers set out to meet each other from two
places 180 miles distant; the first goes 3 mUes an hour, and
the second goes 1 mile more per hour than one-fourth of the
number of hours before they meet. When will they meet?
8. A farmer bought a number of calves, sheep, and pigs,
the number of calves being equal to that of the sheep and
pigs together. For the calves he gave 64s. a head, and for
the sheep twice as many shillings as there were sheep. He
paid £153. 12s. for the calves and sheep together, and £36.
12s. for the pigs — a pig costing as much as a sheep and calf
together. Find the cost of the sheep per head.
9. There are two squares, and an oblong whose sides are
equal to those of the squares, and it is noted that three times
the area of the first square exceeds four times the area of the
oblong by 3 square feet, while twice the area of the square,
together with three times the area of the rectangle, is 36
square feet. Required the sides of the squares.
10. The sum of two quantities is equal to 6 times the
square of their product^ and the sum of their cubes is equal to 36
times the product of their fifth powers. Find the quantities.
11. The solid content of a rectangular parallelopiped is 60
cubic feet, and the total area of the side is 98 square feet,
while the sum of the edges is 48 feet Required the
dimensions.
316 ALQEBRA«
12. The products of the number of units of length in the
sides of a polygon of n sides, .when taken n — 1, together are
respectiyely o^, a^ a^ &c, a^ Bequired the leng^ of the
sides.
13. A, B, and C can together do a piece of work in a day,
and Cs rates of work is the product of the rates of A and B.
Moreover, C is one-fifth as good a workman as A and B
together. Find the respective times required for A, B, C to
do a piece of work.
14. The compound interest of a certain sum of money for
3 years is a, and the third yearns interest is b. Find the
principal and the rate per cent.
15. A owes B £a due m months hence, and also £h due
n months hence. Find the equated time, reckoning interest
at 5 per cent per annum.
16. Find three quantities such that the sum of any two is
equal to the reciprocal of the third.
1 7. Find three magnitudes, when the quotients arising from
dividing the products of every two by the other are respec-
tively a, 6, c,
18. The sum of three quantities is 9, the sum of their pro-
ducts, taken two and two together, is 23, and their continued
product is 15. Show that tike three quantities are the roots
of the equation a' - 9 a' + 23 a - 15 = 0.
CHAPTER in.
Indices.
8. We shall reserve the discussion of 'the complete theory
of Indices for Vol. 11. , confining ourselves here to a few simple
cases, and giving a few examples involving fractional and
negative exponents.
9. Fractional exponents.
Def. — The numerator of a fractional exponent indicates
the power to "whidi t\ie c^«^^\\."^ tsixms^ \3fc raised, and the
INDICES. 317
denominator the root which must be taken of the power so
obtained.
Thus, ai = ila\ a^ = >', aJ = >« =a»;
and generallv a** = va"*.
The above definition is that which follows at once if we
assume the law proved in Art. 24, page 159, viz., d^ x a**
= 0** + •* to be true, whatever be the value of m and n.
Thus we have—
a** J = a** X a** X a" to w factora
— -1. "* 4-^
= a»» n'*'ir-~' = a*^"' = «"*•
::r + '? + r-*o» terms _ ^.« __ _m
Hence, taking the wth root,
a^ = I^aT.
a9j8 = ««• •
Let oj = («»)"• = ^7 ("^^y, ty I>ef., Art. 9.
. of = {a^y = (^^X
or, raising each side to the qth power, we have—
?r
Hence, taking the (5'«)th root, we have a? = o«« .
IL To show that a« x 6 » = (a6)«.
Now,a^ X b^ ='V^ ^■>!/^ = /y«^^= VW^
= (a5)« , by Def!, Art. 9. "'
And so, a? V 6« = (r-)"*
Ex. 1. Multiply together a*6*c4 by a*6tci
Adding together the indices of like letters, we have—
i + i = ^, i + * = T^' 4 + * = H-
Henoeir *b^ npgiured product is aH^c^,
318 algebra:
- x^y^ — asy - aj^y^
Ex. 2. Multiply x + a; V^ + y by a; — a;^y4 + y.
a; + a;^y* + y
as - aj%^ + y
a^ + aj%i + xy
xy
xy + aj^y* + y*
^ + agy + y*
Ex. 3. Divide a; - y by a;» - y^.
jcJ — y^)aj - y(a5* + a3»y» + y«
X - aj^y*
a;»y» - a;*y
a; V - y
^y\ - y
Ex. V.
Find the value of —
1. (a«)^ a*, («-»)-*, (a*)i
2. (a + a;)^(a' + 2 ax + a?)i, (a* - a;*)* (a* + »^)i
Multiply together —
3. a^ - a^aj^ + x^ and a^ + a^x^ + a;*.
4. a;^ - y^ and a?"^ + y~^.
5. aj + ajiyi + y and a?"" - aj^iy""^ + y" .
Divide —
6. a^-6 by a - 6*, a* - 6* by a* - 6*
7. a;^ - fljyi + aj^y - y* by a?^ + «*^*
8. a + 6+c — 3 o^J^c* by a* + 6* + o**
SUBDS. * 319
Show that — •
m
12. (cB" - o™) -f « - a"")
= (^2-+ a^ ((c* + a"^) (««'+ a?) (a?" + a^%
Find the square roots of —
13, a + 6 + c + 2 (a*6^ + a^c^ + fcic*).
U. ixy^ — 12a;% + 17ar^y* - I2x^y^ + 4a:?.
15. d^h-'^ - 4a5-i - 8a-^6* + 4a-% + 8.
Find the cube roots of —
16. a;* + 9a;V + 6a5* - 99ai* - 42aj* + 441aj* - 343.
17. a*y-^ + 3a^y~* + 3ar^y-* + 1.
18. ah{l + 3a-*6J + 3a-56* + a-*6) (aj-^ - 3a?5--'
+ 3a*6-* + 1).
CHAPTER IV.
^UBDS.
12. A surd quantity is one in which the root indicated
cannot be denoted without the use of a fractional index.
Thus, the following quantities are surds : —
320 * ALGEBRA.
Since, from what has been explained in the last chapter,
these quantities may be written thus —
it follows that surds may be dealt with exactly as we deal
with their equivalent expressions with fractional indicea
It is evident that rational quantities may be put in the
form of surds, and conversely, expressions which have the
fomh of surds may sometimes be rational quantities.
. Thus, a* = 4^(^ = il^;
and ila^ + 3a«6 + Zah^ ^r V" = SI {a + hf = a + 6.
18. A mixed qucurvtity may he expressed as a swrd.
Thus, 3 4/5"= ilZ\ ilb= >5'3» X 5 = 4/l357
andso, a5/v^= v^af*. ^y = ^|af^.
14. Conversely, a surd may he expressed as a mixed
qiumtUy, when the root of any factor can be obtained.
Thus, n/18^ = ^/9a«6« x 2a
= ^/9^^ J2T= 3ahj2a.
And iy(arr6^)^= Hj^FT^f^flT^
« ^(a» + hya^y"'^^ = (a' + l^fity ^.
15. i^ractiemo^ «t^c2 expressions may be so expressed that
tlie sv/rd portion mmf he integral.
The process is called rationalisdng the denominator^ and is
worth special notice.
It is much easier to find approximately the value of a/21,
and divide the result by 7, than to find -the values of ^3
and J7, and divide the former by the latter.
Ex 2. Reduce to its simplest form ijj-^ —
/ agy __ l xy{b - c) ^ s/xy{b - c)
suhds. 321
Ex. 3. Find the arithmetical value of
2 - ^/3*
The denominator is the difference of two quantities, one of
hich is a qiwdratic surd.
Now, we know that (2 - ^3) (2 + V^) = 2« - ( ^3)^*
= 4-3 = 1, and hence we see that by multiplying
numerator and denominator by the sum of the quantities
in the denominator we can obtain the denominator in a
rational form,
Thn« — i— = ^(^ + J^) ^ M2+ V 3)
' 2 - V3 (2 - ^/3)■(2 + n/3) 2^ - ( JZf
= 4(2 + 1-73205) = 4(3-73205.)
= 14-92820.
naso ,^ = _^H^srr^^{W
' 4 n/2 + 3 ^/3 (4 ^/2 + 3 n/3) (4 ^/2 - 3 n/3)
^ 4(4^/2'- 3 V3) ^ 4(4 72"-- 3 ^/3) _- .^ 72-3 73)
(4 n/2)« - (3 JZf 32-27 ^ ^*
We shall now give an example when the surds are not
quadratic.
a
Ex 4. Eationalize the denominator of -r r*
05* - y^
Since {x^f^ - (y*)^^ is (Art. 29, page 175) divisible by
a;i - yi, it follows that the rationalizing factor is their
quotient, which is easily foimd.
16. Surds may he redticed to a com/mon index.
Ex 1. Express ^a and yb as surds having a common
index.
Since H^a = a»», and J^h = 6»», it follows that, by
reducing the fractional indices to a common denominator, the
given Burds become respectively a»ft»*, b"*», or Nlar,**y\r*
^ X
322 ALGEBRA.
Ex. 2. Reduce ^^ and ^7?^ to a c5ommon index.
The least common denominator of the fractional indices of
the given surds is 4 x 3 or 12. Hence we proceed as
follows: —
When the student has had a little practice, the first two
steps of each of the operations may be omitted.
17. Addition and subtraction of similar surds.
Def. Similar surds are those which have the same irra-
tional factors.
Ex. 1. Find the sum of Vr2", 5 \/277 - 2 »jTb.
We have —
>/l2 + 5 V27 - 2 >/75
= n/2« X 3 + 6 n/3« X 3 - 2 Jb^ x 3
= 2V5'+6x3V3-2x 5 V3
= 2 V3 + 16 V3 - 10 V3
= (2 + 15 - 10) V3 = 7 Vs;
Ex. 2. Simplify—
'S a^ -- 2ab + b^ " ^ a"" +
- 2a6* + 6»
2a6 + 6^
The given expression —
= / (g + 6)^6 / (g - 6)'^6
^/ (g - 6)3 *" N (g + 6)«
\ a + b^ ^g-6 a + 6^^
a - b
= (« + 6)» - (a- by , A ah ,_
(* -b){a + b)' ^* = ^^r^» ^*-
18. MuUiplicalion and division of swrds.
The following exam^lo^ will best illustrate these opera-
tions ; —
SURDS. 323
Ex. 1. Multiply a \fs?yz by b/Jxt/^u.
"We have, a ija?yz x h Jxfi^u = ah ija?yz x ay'w
= ah ija^yhiz = aboi?i^ Jvz*
Ex. 2. Multiply a sjh + c >/5^by a — VSS.
Arranging as in the case of rational quantities, we have —
a ijh + c tjd
a — tjhd
a^^fi •{- OG Jd
(^a^ - cd) Jb + a(c - b) ^d
Ex. 3. Divide ajbhyb Ja.
We have, *-^ = " ^f ' ^ = iL^ = -J V^
ova b *Ja, sla o.a
When the divisor is a compound quantity it will generally
he the best to express the surds as quantities with fractional
indices, and proceed as in ordinary (^vision.
19.The sqmi/re root of a raiimud qua/atUy camiot be partly
rational and pa/rtly irrational.
If possible, let Ja = m + Jb;
then, squaring, a ^ m] + 2m ^b ■{• b;
or,2mJb = a'- {m^ + b);
or, Jb = tZ-^p^;
that is, an irrational quantity is equal to a rational quantity,
which is absurd.
20. To jmd the aqua/re root of a binomial, one ofwJwse
terms is a quadratic av/rd.
Let a + V6 be the binomial.
Assume tja + ijb = >Jx + Jy, (1);
then, squaring, a + \lb = aj + y + 2fs/xy^ (2).
Equating the rational and irrational "^^Qix:^ (^io^A^V^^
324 ALOEBBA.
fc i- y = a (3.),
and2V^ =*Jb or iicy = b (4.)
From (3) and (4) we easily findo; = \{a + »/a* - b),
fi^d y = i (» - Va"* - b).
Hence, from ( 1 ), the square root required is —
Note.—- It is evident that, unless (a' >- () is a perfect square, our
result is more complicated than the original expression, and therefore
the above method fails in that case.
Ex. 1. Find the square root of 14 + 6^5.
Let Vl4 + 6V5 = aJx + Jy (1.)
Squaring, then, 14 + 6V5 = a? + y + ^*Jxyl
Hence, equating the rational and irrational parts-*.
^ X + y =: 14 (2.),
2>Jxy = 6V5or4iB2/ = 180 (3).
From (2) and (3) we easily find cc = 9, y = 5.
Hence the square root required is V9 + tjb or 3 + a/5.
Ex. 2. Find the square rool^ of 39 + Vl496.
Let n/39 + VU96 = >Jx + 'Jy.
Squaring, &c., we have, a; + y = 39 ;
and 4:xy = 1496.
From these equatiouB we easily find x = 22, y = 17.
Hence, the square root required is V22 + Vl7.
21. The squa/re roots of qucmtitiea of this hind may often
,be found by inspection.
Ex. 1. Find the square root of 19 + 8 Vs.
We shall throw this expression into the form a* + 2 o6
+ b\ which we know is a perfect square.
Dividing the irrational term by 2, we have 4 V3. Now
all we have to do is to break this up into two such factors
that the sum o{ ^e\x: «c]^aAs«% ahall be 19. The factors are
evidently 4 aad *J 3*
suBDS. 329
Thus, we have 19 + Ss/T= (if + 2 (4) Vs + (VS)^
= (4 + V3)l
The square root is therefore 4 + Vs.
Ex. 2. Find the square root of 29 + 12V5r
We have 29 + 12^5"= (3)^ + 2(3)2^/5'+ (2^/5)«
= (3 + 2 ^/k)l
The square root is therefore 3 + 2 t/5.
Ex. VI.
Express with fractional indices —
1. ^/^, ir^b\ if^, v^.
>Jdb ijxy s/a*^
Reduce to entire surds —
3. 3 >/3, 4 Ji, I n/15, 3 v^f
4. 4.2^,9.3"*, 4.2"* !(!)'"*.
5. 3Va^, a J-, (a + x) J /^ " ^ .
Reduce to a common index —
6. v2, v3. 7. v2, v3.
8. 2 ^2, 3 -ys. 9. v^^ v^g:
10. (a + aj)*> va - X. 11. oi* «> hp «•
Simplify—^
12. n/12, 4^48, 3 -s/28, ^ v^648.
13. V4a« + 4a«6, il^^¥TT\ ^/^^^^
'V 64 a
, , n 2^ : 'p 3-^2 -
326 hiiimmk
Find tiie Tsfaie
h
18. 4^27ir^«ft» - 4^8d*-+» + 3 ^'STi^.
Multiply—
19. a + \/ar+ hhj Ja " Jb^td + bihj Ja ^ >Jb,
20. (« + y)i by (x + y)4, « + 6N/?by a" •- ^ Jd ^ VI
22. a* + fti + cl + cZi by ai - fti + ci - d*.
Divide —
23. ai* + ary + ^bya: + x^y^ + y.
24. a? - y^bya;4 + yl.
Bationalize the denomixiafcarB of —
3 • 4 1
26.
2 + VS' 3 n/2 - 2 V3' ^5 - V3
27 5 ^ ^ 'Jl±Jl,
* 1 + ^/2 + V3' >/2 + V 3 + n/5' ^3 - n/2
28- a;i . yi' ^ + VS' aj + a?iy* + y
Find the square roots of —
29. 11 + 4 V7, 8 + 2 ^15, 30 - 10 Jf.
30. 8 + 2 VT2, 9 - 6 V2) 20 - 10 ^/3;
BATio. 327
CHAPTER V.
BATIO AND PBOPOBTION.
Batio.
22. The student is referred to Chapter II. of the Arith-
metic section of this work for definitions and observations
which need not be repeated here.
23. A ratio qfgrecUer ineqiuilU^ ia diminished, and a ratio
o/less inequality ia increased, hy iner easing the terms of the
ratio hy the same qua/atity.
Let a : 6 or r- be the ratio, and let each of its terms bo
increased by m. It will then become -r *
Nowj-^^ g~ as (rt + w)6 5 (i + «>»)flf>
-h m ^ ' ^ '
or, as a6 + &m S a5 + ami] or, as 5m S am, or as i S cti»
Hence the ratio -r Is increased when h ^ a, that is, wheii
it is a ratio of less inequality; and is diminished when
6 < a, that is, when it is a ratio of greater inequality.
Cob. It may be shown in the samel i^ray that —
A ratio of greater inequaMy is increased, and a ratio of less
inequality is diminished, by diminishing 1M terma of the ratid
by tJie scmie quantity,
24. When the difference between the antecedent and con-
sequent is small compared with either, the ratio of the higher
powers of the terms- is found by doubling, trebling, &c, tiieir
difference.
■ ■ — X
' ^ tfX -I- S/* J
iLet a + a; : a or be the ratio, where x is small
a
compared with a.
Then "^ ,— ^ = r^ « 1 + — nearly =
a + 2aj' , '
nearly.
328 ALGEBRA.
(a + xf oP + 3 a'a + 3 oo? + a? , 3 as ,
^^ i-^ = -j = 1 + — nearly =
or or a ^
t% •- •— • •■-- —
a + ox - -
nearly; and so on.
Ex. (1002)« : (1000)^ = 1004 : 1000 nearly.
(1002)» : (1000)» = 1006 : 1000 nearly.
Proportion.
26. Proportion, as has been already said, is the relation
of equality ezpreased between ratioa
Thus, the expression aih = c\d^
or a:b :: c:d,
is called a proportion.
S6. The following results are easily obtained : —
.) Smce r = j> ^^ rX~ = jX-or-=j,
• a e d c e d
.*. a:c :: b:d (aUermmdo).
^ ' 6 d a c
.'. b: a :: d: c (mvertendo).
Also, by Art 64, page 214, we have —
(3.) a + b:b :: c + d: d {cornponevido).
(4.) a " b:b :: c •- d: d (dividendo).
(5.) a •- b:a :: c •- d:c (converiendo).
(6.) a -^ b : a - b :: c + d : c - d (eomponendo and
dividendo).
87. Jfaib:: c:d and e :/:: g : h, we may compound
the proportions.
Thus we have V = r (l),and^=^ (2).
(l)x(2);then,|=|.
PROPORTION. 329
And in tte same way we may show that, if the correspond-
ing terms of any number of proportions be multiplied together,
the products will be proportional.
28. If three qucmtities a/re in continued proportion, iihe first
has to the third the duplicate ratio of what it has to the
second.
Let a, 5, c be the given quantities in continued propor-
tion; then —
a _ h
I" c
.rr- a a b a a a^
Hence, — or =- x - = i Xt = £5 9
' c c ¥
.'. a : c : : a^ : h\
And, similarly, if a, h, c, d are four quantities in continued
proportion, a : d :: a^ : b% that is —
The first has to the fourth the triplicate ratio of what it
has to the second; and so on, for any number of quantities.
29. We shall now give one or two examples of problems
in Proportion.
Ex. 1. If a: bi: Old, prove tha t li , ^ = ( r — 3) •
Let r = ^ = oj; .•. a = bx, and c = dx.
Hence, ^J^ = (^)' > i^)' = ^1±^ , then, after-
^ ' (« + c)» {bx + dxy (6 + df ^
Ex. 2. If a :b :: c :d, prove that 7 7=.^ j=^.
a c
Let 7- = -- =«:.•.« = fee, and c = c2a;.
6 a
HencC)
a - b bx - b " X - I JbdT.x - V^
isjbx.dx + V^ci _ Vac + fjbd
Jbx.dx ^ sIM Joe- ivIbdH
330 ALGEBRA.
Or, it may be worked thus —
Since J = |,wehaye J| = J|,
Hence^ by Art 26 —
a + h _ Jac + sfbd
a - 6 Wac - n/65'
Ex. 3. If a : 6 :: c:d :: e :/, show that
a __ imoT + nc** + pe^y
^* I = 1=7=" • ^^^-
a** _ c** _ ^ ^ r^
(f = c^*'af, /. TMJ** = TK^'af >, and .*. by addition,
wtoT + wf
.'. Equating (1) and (2), we have—
i
b \mb^ + ndr + pf'J
Ex. VII;
1. Compare the ratios a -{• b:a ^ b, and a^ + b^ia^ - 6%
2. Which is the greater of the ratios a + 6 : 2 a, and
26 ; a + 61
3. What quantity mxiat \j^ sviUraxjted from the consequent
fef the ratio a : 6 in oxdet to TO»a5ft\t ««;j\ai\<^^^\»J^v^ c -M
PKOPORTION. 331
4. Compound the ratios l-cc^il+.y, aj-ajy^rl+ar*,
and 1 : 05 - 05*.
5. There are two numbers in the ratio of 6 : 7, but if 10 be
added to each they are in the ratio of 8 : 9. Find the
numbers.
6. In what cases iax +— =*or-<51
X
7. If = = , show that a + b + c = 0.
X — y y — z «-05'
8. Find the value of 05 when the ratio 05+ 2a:x + 2 his
the duplicate ratio of2a; + a + c:2o5+ h + e,
9. Find 05 when the ratio x -- h : x + 2a — 6is the
triplicate ratio ofaj-a:fl5 + a-6.
,^-r-«J + v y + z 05 + «- , 4i.,-
10. If — — y = J-—, — = — ; — , show that each of the frac-
a + oo + cc + a'
tions IS equal to r—. — , and that — = i = ""•
^ a + 6 + c' a c
^^- ^h = ;^ = />^en eax)h la equivalent to ^^ ^ ^^ ^ ^y,
hence, show that —
a h c
2z + 2x — y ~ 2x + 2y ^ z ^ 2y + 2z^x
, 0^ 2/ ^ .
^^®^ 2a + 26-c'" 26 + 2o^a ~ 2c + 2a- b'
12. if a : 6 : : c: d, then
a ^ b : c + d :: si a? + a6 + 6* : Vc* + C(/ + cf.
13. Find a fourth proportional to the quantities—
o; + 1 a? -¥ X -{- 1 05^+1
05-1' 05^-05+1 05*— 1*
14; Find c in terms of a and b "When —
(1.) a \a \\a - 6:6 - c.
(2.) at6::a — 6:6-c.
(3f) a : c ; : a - b ; b - c.
332 ALGEBRA.
15. If a, h, c Bjce in contmued proportioii, show tliat
16. 1£ a:b ::c :d, then —
Va«* + 6'* : Vc** + cP* : : (a - ft)* : (c - d)\
17. From a vessel containing a cubic inches of hydrogen
gas, b cubic inches are withdrawn, the vessel being filled up
with oxygen at the same pressure. Show that if this opera-
tion be repeated n times successively, the quantitv of hydro-
(a - 6)*
gen remaining in the vessel is — ^jur- cubic inches, when re-
duced to the original pressure.
18. If, in Ex. 34, page 225, (oi, a,, a,), (^, ^jy ^)) ^^^^
(ci, C], C3) are corresponding terms respectively, show that
SECTION III.
PLANE TEIGONOMETRY.
CHAPTER I.
MODES OF MEASUBINO ANGLES BT DEGREES AND GRADES.
1. We are able to determine geometrically a right angle,
and it might therefore be taken as the unit of angular mea-
surement. Fracticallj, however, it is too large, and so we
take a determinate part of a right angle as a standard.
In England we divide a right angle into 90 equal parts,
called degrees, and we further subdivide a degree into 60
equal parts, called mimdes, and again a mmvJte into 60 equal
parts, called seconds. This is the English or sexagesimal
method.
In France the right angle is divided into 100 equal parts,
called grades, a grade into a hundred equal parts, called
minutes, and a minute into 100 equal parts,' called seconds.
This is the French or centesimal method, and its advantages
are those of the metric system generally.
The symbols ^ ', ", are used to express English degrees,
minutes, seconds respectively, and the symbols ', \ ^\ to express
French grades, minutes, seconds respectively.
Conversion of English and French Units.
2. Let D = the number of degrees in an angle,
and G = the number of grades in the same angle ;
then ^expresses the angle in terms of a right angle;
and so also does ^qq*
334 PLANE TBIGONOXETRT.
D G D G
Hence, ^^ - Jqq or-^ - j^.
. D = ^G = G --G (1).
andG = ^D = D + |d (2).
Hence the following rules : —
1. To convert grades into degrees.
From the number of grades suBTBAcr •j'^, and the remaiii*
der is the number of d^rees.
2. To convert degrees into grades.
To the number of degrees add ^y and the sum is the num-
ber of grades.
Ex. 1. Convert 13^ 18* 75'' into English measure.
No. of grades = 13-1875
Subtract ^^ of this = 1-31875
.-. No of degrees = 11-86875
60
52/12500
60
7''-600
Ans. ir 52' 7"-5.
Ex. 2. Convert 18* 7' 30'' into French measure.
No. of degrees = 18-125
Add i of this = 2-013S
.-. No. of grades = 20-1388
Ans. 20^ 13' 88''-8.
3. An angle may be conceived to be generated by the
revolution of a line about a fixed point. T^us —
Let OA be an initial line, and let a line, OF, starting
from OA, revolve mt\i O ^ oentre^ and take up sucoes-
siyelj the j)OBitiona OP^ OS^, O^^, O^^
MODES OF M£ASUBIXG ANGLES. 335
Now the magnitude of an angle may be measured by
the amount ofVwming required to generate it. When, there-
fore, the revolving line reaches the position OB, we may con-
ceive an angle to have been
generated whose magnitude ^
is two right angles. And,
further, when the revolving
line assumes the positions '^|
0P„ OP4, the angles AOP„
AOP4 (the letters being read
in the direction of revolu-
tion) are angles whose mag-
nitudes are each greater than two right angles. Indeed,
when the revolving line again reaches the position OP, we
may conceive an angle to have been generated whose magni-
tude is four right angles. Lastly, tf the revolution of the
line OP be continu^, we may conceive of angles being
generated to whose magnitude there is no limit.
Ex. I.
1. Express 39' 22' 30" in French measure, and IS*' 15'
75^' in English measure.
2. One of the angles at the base of an isosceles triangle is
50**. Express the vertical angle in grades.
3. Divide an angle of n degrees into two such parts that the
number of degr^s in one part may be twice the number of
grades in the other.
4. Two angles of a triangle are respectively a^ ^, express
the other angle in degrees and grades.
5. If f of a right angle be the unit of measurement, ex-
press an angle which contains 22*5 degrees.
6. Show how to reduce English seconds to French seconds.
7. K the unit of measurement be 8**, what is the value
of 10^.
8. If two of the angles of a triangle be expressed in grades,
and the third in degrees, they are respectively as the nimi-
bers 5, 15, 18. Find the angles.
9. What is the value in degrees aad ^di^^cjl «sl%x\2^<^
836
PLANE T^IGONOMETRT.
winch is the result of the revolution of a Hne 3^ times
round.
10. In what quadrants are the following angles found :—
145^ 96^, 327^ 2.72', 272°.
11. If a<* be taken as the unit of angular measurement,
express an angle containing 5^.
12. What is the unit of measuren^ent when a expresses-
of a right angle ?
CHAPTER n.
THE GONIOMETBIC FUNCTIONS.
4. It was formerly usual in works on Trigonometry to give the
following definitions : —
Let a circle be described from centre A,
^ with nbdius AB supposed to be unity, then—
(1.) The sine of an arc BC is the perpea-
dicular from one extremity, O, of the arc
upon the diameter passing tlurough the other
extremi^B.
Thus CSisthe sine of the are BC.
(2.) The cosine of an arc is ^e gineoftk
complement of the arc.
Thus, since DO is the complement of BC,
S^ is the cosiNX o/tfte arc BC,
(3.) The tangent of an arc BC is a line
drawn from one extremity, B, of the arc
touching the circle, and terminated in the diameter which passes
through the other extremity, 0, of the arc.
Thus, BT M ^ TANGENT of the arc BO,
(4.) The cotangent of an arc is the tangent of tlie complement of
the arc.
Thus, DT' is the cotangent of the arc BC,
(5. ) The secant of an arc BG is a line drawn from the centre through
one extremity, 0, of the arc, and terminated in tiie tangent at the
other extremity.
Thus, AT is the secant of the arc BC.
(6. ) The cosecant oi an arc \b ^«i aecaul of tde complement of the arc.
Thusy AT' i« tht cosECJjar oj tKe arc BC.
TRIGONOMETRICAL RATIOS. 337
(7.) The versed sine is that portion of the radius upon which the
sine falls, which is included between the sine and the extremity of
the arc.
Thus, SB is the versed sob of the arc BC.
(8.) The coversed sine is the versed sine of the complement of the
arc.
Thus, S1>ist7ie coversed sjnTEoftlie arc BC,
(9.) The suversed sine is the versed sine of the supplement of the arc.
Thus, 'R'^ is the suversed sine of ilie arc BO,
Representing the arc BO by A, it is usual to write the above func-
tions thus : — Sin A, cos A, tan A, cot A, sec A, coflec A, vers A,
covers A, suvers A.
By mere inspection, the student will see that the following rela-
tions hold ; —
(1.) Sin A = CS = ?? = ?| = -^ ^
(2.) CosA = S'O
1 AS AT' cosecA
AS^ AS _ AB, 1
1 AC AT '^ sec A'
]i\ Tan A - T^T - ^'^ - ^T ^ AD _ 1
(3.) Tan A = BT p: -p - -^ =:= ^^, - j^^.
(4.) Sin« A + cos« A = CS« + S/C« = CS« + AS* = AC« = 1.
(5.) Sec' A = AT« = AB« + BT« = 1 + tan« A.
(6.) Cosec' A = AT/« = AD« + DT/« = 1 + cot' A.
(7.) Tan A = BT = IB "" AS "" S^ " ^^STa'
(8.) Vers A = SB = AB - AS = AB - S'O = 1 - cos A.
(9.) Covers A = ST) = AD - AS' = AD - CS = 1 - sin A.
(10.) Suvers A = B'S = B'A + AS = 1 + cos A.
It is more convenient, however, to define the sine, cosine, &c., as
in the next article, according to which definitions they are commonly
called Trioonometrical £^tios. The student will see that if the
above definitions be so far modified that, instead of the Imes^ them-
selves, the ^niometrio functions be taken as the ratios which the
lines respectively bear to tiie radius, they are included in the defini«
tions of the next article.
Trigonometrical Ratios.
6. Let BAG be any angle, which we may denote by A,
and P any point in the line AC. Draw PM perpeftdicul«c
toAB.
^ Y
338
PLANE TBIGONOMETRY.
Then (1.) Sin A = jgg^'^dicular ^ m
^ ' hyp. AP
_ . ^ . base AM
2.) Cos A = 5^ = -^.
^^^ ^ perp^dicdar ^ PM
4.) Cot A =
base
base
AM
AM
perpendicular PM*
5.) Sec A = JyP- = ^.
' base AM
6.) Cosec A = j.' , — = -^^jTjt-
' perpendicular 'PM
7.) The 'oefrsed sine is the remainder after subtract-
ing the cosine from unity, or —
vers A = 1 - cos A.
8.) The coversed sine is the remainder pfter subtract'
ing the sine from unity, or —
covers A = 1 — sin A,
[9.) The suversed sine is the sum of the cosine aud
imity, or —
Buvers A = 1 + cos A.
The last three are not much used in practice.
6. Compaiing (1.) and (6.), (2.) and (5.), (3.) and (4), of
the last article, we see at once that the sine and cosecant^
the cosine and secant^ and the tcmgent and cota/ngent, are
reispectively eact t\i"ft TecVgtoca)^ c>i^^ ^'^^^
TRIGONOMETBICAIi RATIOS. 339
We therefore have —
(1.) Sin A = , cosec A = -r
cosec A sin A
(2.) Cos A = jTi sec A = r .
^ ' sec A cos A
(3.) Tan A = -~^, cot A = r-^-
^ ' cot A tan A
Further —
(4.) Srn^ A + cos« A = -j^ + j^ = j^,
AF
~ AP2 " ^•
Hence also, transposing and taking the square root —
(5.) Sin A = ^/l - cos^ A.
(6.) Cos A = VI - sin^ A.
And again —
,^, ^ ,. AP2 AM^ + PM^ wI'M' 1 ^+ 2 A
(7.) Sec»A =-^,= ^^, :. H. _,= 1 + tan^A.
/ox ri , A ^I"' I'M' + A^' 1 AM^
(8.) Cosec A = y^ = pjja = 1 + -pjp
= 1 + cot^ A.
,^ , ^ , PM PM AM . . .
(9.) Tan A = -^2i = AP "^ AP = sin A -f cos A
_ sin A
~~ cos A"
,,^, ^ ^ , AM AM PM . . .
(10.) Cot A = p^ = -^ -f -^ = cos A -r sin A
_ cos A
~ sin A*
The student must make himself thorou^ly master of ik^
results in this article.
340 PLANE TBIOONOMETBT.
7. To express the trigonoaietrical ratios in terms of the
sine,
(1.) Cos A = VI - sin« A, by Art 6 (6.)
(2.) Tan A = ^^^, by Art 6 (9.),
_ sin A
(3.) Cot A = ?^, by Art 6 (10.),
^ ' sin A
^ Jl - sin' A
"" sin A '
(4.) Sec A = --^x' by Art 6 (2.),
^ ' cos A
- 1
Vl -sin'A*
(6.)CosecA = -5l^,byArt.6(l.)
sin A
Ex. K sin A = I, find the othelr trigonometrical ratios.
We have, cos A = >/l -(i)« = Vl - A = I-
sin A
8. To express the trigonometrical ratios in terms of ike
cotangent
(1.) Sin A = — ^ = .. ^ ^.■ . by Art. 6 (8.)
cosec A V 1 + cot' A
(2.) Cos A = ~ =—==!==., by Art 6 (7.)
sec A VI + tan'A ^
1 cot A
/s/i +
1 Vcot'A + l'
oot'A
(3.)TmiA.= ^
TRIOONOUETSICAI. BATIOS. 341
(4.) Sec A = — ^- = 1 T-,=^A=^, by (2.) above,
COS A V cot' A + 1
_ Vcot' A + 1
"" cot A
(5.) Cosec A = VI + cot' A, by Art. 6 (8.)
And in the same way the trigonometrical ratios may be
expressed in terms of any one of them.
Ex. II.
1. Given sin A = -Jf, find the other trigonometrical
ratios.
2. Given tan A = f|^ find the remaining trigonometrical
ratios.
3. If cot A = a, show that sin A = ■ .
4. If vers A = 6, then tan A = ^ — ^ — .
1—6
5. Construct by scale and compass an angle (1.) whose
cosine is f ; (2.) whose tangent la ^; (3.) whose secant is »J2;
(4.) whose cotangent is 2 + ^/3I
Prove the following identities : —
6. (Sin A + cos A)' + (sin A - cos A)' = 2.
7. Sec' A + cos' B . cosec' B = cosec' B + sin' A. sec' A.
8. Sec' A . cosec' A = sec' A + cosec' A.
9. Sec A . cosec A = tan A + cot A
10. Sin^A - cos^A = (sia' A - cos' A) (sin* A + cos* A).
|, SecA + tanA _ cosec B - cot B
Cosec B + cot B sec A — tan A
. . . sin* A + sin' A cos' A + cos* A
12. 1 + sm A cos A = z : — r r •
1 - sm A cos A
13. (arcosa + ysin0)(a:sme + ^^coseji - (ojcose - ysine)
(xmnd — pco»t)) - 2ay.
342
PLANE TBIOONOMETBY.
14. (asinocost^ + rcosOcos<^) (6sin6sin(^+ rsmooos^)
— (6 sin coS(^ - r sin sin <^) (a sino sin 4$ + r cos sm^i)
= r sin o(r cose + a sine).
15. If JB = r sin 6 cos <t>, y = r sin sin <^, z = r cos^, show
that or + y* + a* = r*.
16. If a = 5 cos C + c cos B, 5 = a cos C + c cos A,
c = a cos B + 6 cos A, show that a^ + 6*-6*= 2a6cosC.
17. Given sin" A + 3sin A = ^, find sin A.
18. Given cos" A — sin A = ^, find cos A.
19. Solve sin A — cos A = -j=, for sin A.
20. Find tan A, when tan A + 1 = a/j^ sec A.
21. Given a cos A = 6 sin A + a, find cot A.
22. Given tan' A - 7tan A + 6 = 0, find tan A.
23. Show that Vl + 2 sin A cos A + Jl — 2 sin A cos A
= 2 cos A or 2 sin A, according as A is between 0° and 45°, or
between 45° and 90°.
24. Given m sin" A + n sin" B = a cos" A,
m cos" A + 91 cos" B = 6 sin" A^ find sin A and sin B.
CHAPTER III.
CONTRARIETY OP SIGNS. — CHANGES OF MAGNITUDE AND SIGN OF
THE TRIGONOMETRICAL RATIOS THROUGH THE FOUR QUADRANTS.
9. We have explained at some length the meaning and
[ use of the signs + and - in
algebra. They have a similar
interpretation in trigonometry.
1. Lines. — Draw the horizontal
^ line A A, and draw BB' at right
angles, meeting it in O. Then
considering O as origin,
(1.) All lines drawn to the ri^A<
IB parallel to A 'A are called positive,
CONTRARIETY OP SIGNS.
343
and all lines drawn to the left parallel to A'A are called
T^ative,
(2.) All lines drawn upwa^'ds parallel to B'B are called
positive, and all lines drawn dovmwa/rds parallel to B 'B are
called negative.
(3.) lines drawn in every other direction are considered
positive, as is therefore the
revolving line by which angles
may be conceived to be gen-
erated.
2. Angles, — ^A similar con-
vention is made forangles. Let
OA be an initial line, and
let a revolving line about the
centre O take up the positions
OP and op;. Then—
(1.) That direction of revolution is considered positive
which is contra/ry to that of the hands of a watch, and the
angle generated is a positive angle.
(The positive direction is then upwa/rds,)
Thus, AOP is a positive angle.
(2.) The negative direction of revolution is the same as that
of the hands of a watch, and the angle thus generated is a
negative angle.
(The negative direction is then doumwa/rds,)
Thus, AOF is a ne-
gative angle.
Hence, if the angles
AOP and AOP' be
'of the same magni-
tude, and
We have —
ZAOP = A,
thenZAOP'= - A,
10. We will now
examine the trigono-
metrical ratios for
angles greater than a
right angle^ ami for negative angles.
344 PLANE TRIOOKOMETBY.
Let OPi, OPg, OP3, OP^ represent the position of the Wt
Yolving Ime at any period of revolution in the several
quadrants respectively,
And let PiNj, PoNg, P3N3, l^^N^ be the respective per-
pendiculars from the end of the revolving line upon the
initial line.
Then PiN^, PjNg, P3N3, T^^ are respectively the per-
pendiculars corresponding to the angles generated.
Also, ONj, ONg, ON3, ON4 are respectively the bases of
the right-angled triangles with respect to ike angles in
question.
We have then in the second quadrant —
Sin AOP„ = ?&, cos AOP, = —2
^ OP2 ^ OP2
tanAOPg, = ^-^&c
UJN2
It is therefore evident that the relations between the trigo-
nometrical ratios, which were proved to exist in Art 7,
also hold for angles in the second quadrant — ^that is, angles
between 90** aud 180^
And in the same way we may show that they hold for
angles in the third, fourth, or any quadrant
And again, if we suppose the line to revolve in a negative
direction, and take the position OP ', we shall have P 'Sf the
perpendicular corresponding to the negative angle AOP', and
ON ' the base.
Hence, sin AOP' = ^, cos AOP/ = ^^^,
P'N' \
And the relations proved in Art. 7 may be also similarly
proved to exist here.
Hence the relations proved in Art. 7 hold for any angles
whatever.
Changes of Magnitude and Sign of the Trigonometrical
Batios.
11, Let OP^, OP^^ OP^, OP^ be positions of the revolving
line in the BeveraY c\vLaAr«si\a ic«aj^^Yi^^ \ P^?!^\!i ^^v
CHANGES OF MAGNITUDE AND SIGN. 345
P3N3, P4N4, the respective perpendiculars; and ON^, ONg,
ONg, ON^f the hoses of the corresponding right-angled
triangles.
Then—
(1.) In the first quadrant —
Sin AOPi = ?^^\ cos AOP, ^ ^^\
tanAOPi = '^, &c.
At the commencement of the motion of the revolving line,
the angle AOPj = 0° ;
Also, the perpendicular P^Ni = 0,
And the base ON = OPj.
Hence, we have —
*SinO-^.0,cosO-§| = l,
tan 0» = ^^, = 0.
As the revolving line moves from OA towards OB, Pi^j
increases and ONj diminishes ; and when it arrives at OB,
wehavePjNi = OP^andONj = 0. But the angle generated
is now a right angle. Hence we have —
Sin 90° = ^?i = 1, cos 90° = ~=h
tan 90° = ^ ■- 00.
Hence, as the angle increases from 0° to 90° —
The sine changes in magnitude from to 1 and is + .
The cosine changes in magnitude from 1 to and is + .
The tangent changes in magnitude from to 00 and is + .
(2.) In the second quadrant —
Here the perpendicuh/r l^^c^^ is + ,
and the base ONg is - .
* The student onght properly to look upon the values 0, 1, here
obtamed as the limiting values of the ame, cosine, and Umqent re&\|«iC-
tively, when the angle is indefinitely diminiahed.
346 PLANE TBIGONOMETRT.
Hence the sine during the second quadrant is +, the
cosine is - , and the ta/ngent is - .
Again, as the revolving line moves fix)m OB to OA', the
perpendicular P2N2 diminishes until it becomes zero. Also,
the base ONg increases in magnitude, until it finally coincides
with OA', and /. = — OP^. But the angle now described
is 180°.
Hence we have —
Sinl80'=^^=0,cosl80--g^=-l.
tan 180° = - --^ = 0, &c.
v/JL 2
Hence in the second quadrant — •
The sine changes in magnitude from 1 to 0, and is positive.
The cosine changes in magnitude from to 1, and is negative.
The tangent changes in magnitude from 00 to 0, and is negative.
And in the same way may we trace the changes of magni-
tude and sign in the third and fourth quadrants.
Thus we shall find —
(3.) In the third quadrani —
The sins changes in magnitude from to 1, and is negative.
The cosine changes in magnitude from 1 to Q, and is negative.
The ta/ngent changes in magnitude from to 00, and is positive.
(4.) In the fourth quadrant — '
The sine changes in magnitude from 1 to 0, and is negative.
The cosine changes in magnitude from to 1, and is positive.
The tangent changes in magnitude from 00 to 0,and is negative.
Moreover, as the cosecant, secant, and cotangent are
respectively the reciprocals of the sine, cosine, and tangent, it
follows that their sign^ will follow respectively the latter,
and that their magnitudes will be their reciprocals.
CHAPTER IV.
TRIGONOMETRICAL RATIOS CONTINUED. ARITHMETICAL VALUES
OP THE TRIGONOMETRICAL RATIOS OP 30°, 45°, 60°, «fec.
12. To pnroe that aiii A = cos (90° — A), and that
TRIGONOMETRICAL RATIOS.
347
Using the same figure as in Art. 5, we have-
PM
Sin A = '-prr' = cos APM.
AP
But Z APM = 90^ - A,
.-. SinA = cos (90'' - A),
and similarly —
cos A = sin (90° - A),
tan A = cot (90«» - A),
cot A = tan (90° - A),
sec A = cosec (90" — A), j^
cosec A = sec (90° - A).
13. Ratios of 45°.
In the last figure, suppose Z PAM = 45% then also
Z APM = 90^ - 45' = 45°. And hence Z PAM = Z APM,
and .-. PM = AM (Euc. L, 6).
Hence, also —
AP or Jaw ■¥ PM^ = V2AM2or ^/2TSP.
.-. AP = AM sjlov PM >/2r
Hence we have —
Sin 45'' = sin PAM =:
Art. 8.
Tan 45* = tan PAM =
Sec 45^ = sec PAM =
PM PM 1 .., ,
-jPo = ^,, » ;- = —r- = COS 4o , by
AP PM>s/2 n/2 ' ^
PM PM
- 1 = cot 45", by Ai-t 8.
AM PM
AP _ AM ^/2" ._
" am"" = V 2 = cosec 45^
AM
by Art. 8.
14. Ratios of 30° and 60°.
In the same diagram, suppose Z PAM = 30°, then
Z APM = 90° - 30^* = 60".
Hence, if we conceive another triangle equal in every
respect to APM to be described on the other side of AM, the
whole would form an equilateral triangle whose side is AP.
Hence, PM = J-AP.
348
PLAim TBtOOKOUfiTtl¥.
Now AM = VaF - PM>, .-. AM = ^/Ai»* - (iAP)*
We hence have —
PM iAP
Sin 30- = BinPAM = ^p = ^ = i = cos 60°, by Art 8.
Cos 30' = cos PAM r=
AM
^AP r
-A. =^ = sin 60', by
Art. 8.
AP
Tan 30' =
PM
AM
AP
^^— = JL = J V3" = cot 60', by
v3^p V3
Art. 8.
15. ^0 «/aom> eAa« sin (180' - A) = sin A,
cos (180' - A) = - cos A,
Let Z AOPi = A,
And let the revol-
ving line describe an
angle AOPg = 180'
- A*
Then ZA'OPg =
180° - (180" - A) = A;
Hence, Z AOP^ = A 'OP2.
Hence, also (Euc. I., 26), if Pj^j, PgNg be drawn perpen-
dicular to AA^ P^Ni = P2N2, ON2 = - ONi,
We have therefore —
Sin (180' - A) = sin AOPo = 5^ = -J^i = sin A.
^ ' ^ OPg OPi
Cos (180' - A) = cos AOP2 =-^ = --^1 = - cos A
OP
2
OPi
And similarly —
Tan (180' - A) = - tan A, cot (180' - A) = - cot A.
Sec (180' - A) = - a^<i A.^ QO^ea^^ - ^ ^ oosec A.
TRIGONOMETRICAL RATIOS.
349
16. To show that sin (180° + A) = - sin A,
cos (180° + A) = - cos A,
Let AOPi = A,
And let the revolving
line take a position such
that P^Pg is a straight
line.
Then, evidently, Z AOPo
= 180° + A.
Then, as in last Article —
P3N3 = - PjNi,
ONs = - ONj.
Hence —
Sin (180° + A) = sin AOP3 = ^1 = - -5^ = - sin A.
Cos (180° + A) = cos AOP, = ^= - ^ = - cos A.
And similarly —
Tan (180° + A) = tan A, cot (180° + A) = cot A.
Sec (180° + A) = - sec A,cosec (180° + A) = - cosecA.
17. To show that sin ( - A) = — sin A, and
cos ( -
Let Z AOP = A,
A) = cos A,
And let the revolving line
describe an angle AOP'
= -A.
Then evidently, if PNP '
be drawn perpendicular to
OA, we have (Euc. I., 26)
P'N = - FN".
Hence —
Sin(- A) = sin AOP' =
Co8(- A) = cos AOP' =
P^
op/
ON
op/
= — sin A.
OP '
ON
OP
= cos A.
And similarly —
Tan ( - A) = - tan A, cot (- A) = — c«t A.^
iSw ( -* A) ^ sec A, coaec i^— K^ = - ^'sr^ fe^
350
PLANE TRIGONOMETRY.
Although the results of Arts. 12, 15, 16, 17 have teen
x)yed from diagrams where A is less than a right angle, the
student will have no diffi-
iC culty, if he has understood
the proofs, in deducing the
^ same results for angles of
any magnitude whatever.
18. To show tliat tan ^-
_ 1 - cos A
sin A '
A LetZAOC = A;
Bisect it by the straight line OB, so that Z AOB =
and draw CD perpendicular to OA, meeting OB in E.
(1).
OD ED
2'
Then tan ^^ = tan EOD = ^
Z uD
Now(Euc.VI.,2),g^ = g, and .-. ^^-^ -^,
or
ED
CD
^ CD(OC - OD) ^ CD(OC - OD)
OD ~ OC + OD 0C» -' OD^ CD*
OC - OD OC - OC cos A 1 - cos A
CD
OCsinA
sin A
Q.E.D.
CoR 1. Hence, squaring —
rj,^^, A ^ (1 - cos A)' _ (1 - cos A)<
sin* A ■'I - cos* A
2
.'. Art. 64, page 215,
1 - tan'' -Q-
1 + tan* -o
1 - cos A
1 + cos A'
cos A
___^ .
1
.-. Cos A = £.
1 + \,M:f ^
(1).
TRIGONOMETRICAL RATIOS. 351
Bin^ 9-
1 J A
cos^ ~ cos^ --- - sin^
2t 2i
A
2
sin^ -^ cos^ ij- + flin^
A
cos-^
A
2
A A
= cos^-^ — sin®-^
1 1 • • • 1
■ ••..^J.)
= co8'|:-(l-cos"^)
= 2 008"^ - 1
« . • •(«^.)
c\ / 1 • ^\ 1 1 fl
1 • «
» A /i \
19, To find the trigonomeiHcal ratios of 15% 75°, 120%
135°, 150°.
(1.) SaHos of 120°, _
Sin 120° = sin (180° - 120°) = sin 60° = ^-J,
Cos 120° == - cos (180° - 120°) = - cos 60° = - ^
Tan 120° = - tan(180° - 120°) = - taQ60° = - n/3,&c,
(2.) Batios of I6(f,
Sin 150° = sin (180° - 150°) = sin 30° =i,
Cos 150° =- ^ cos (180° - 150°) =: - cos 30° = - ^,
Tan 150° = - tan(180° - 150*) = - tanSO = - -1.^
Sic.
352 PLANE TRIGONOMETRY.
(3.) Ratios of 15\'
By last Art., tan -- = — ~ — : put A = 30\ or -- = 15°,
2 sin A 2
1 - ^
then tm 15'^ = Iz.^' = J_A ^ 2 _ ^3
sin 30 1 _^
T
From this result we easily get, Art 8,
Sin 15^ = ^^-J cos 15^ = ^^t \ &c.
2 V2 ' - 2 \/2 '
(4.) ii?a«io« 0/ 75'. _
We have, sinys** = sin (90' - 16') = cos 15' = V^ ^-\
2 j^2
cos 75' = sin (90" - m = sin 15' = "^ '\
tan 75' = ta» (90* - IS'') = cot 15' = — ^^
2 — V"
= 2 + ^/3; (fee.
(5.) Eatio8qfU6\
We have, sinl35'= sin(180'- 135°) = sin 45'*= -1=,
v2
cos 135' t= -cos (180' - 135°) = - cos 45'
"" "" n/2'
tan 135' = - tan (180' - 135*) = - tan 45'
= — 1, &c.
Ex. III.
1. Define a negative angle, and show that tan (—A)
= - tan A, when A lies between - 90' and - 180*.
2. Trace the changes of sign of sin A . cos A througli
the four quadrants.
3. Trace tiie chaa^e.^ oi ^\^ c^C cos A + sin A, and of
COS A - wu A>, aa A.vJia»2a^<i^^wa. - ^ \fi^"S5?.
TRIGOXOMETBICAL RATIOS. 353
4. Assuming generally that cos 2 A = cos* A - sin' A,
trace the changes of sign, of cos 2 A as A changes from - 45^
to 315^
5. Write down the sines of 210", 165^ 240% - 120^
6. Show that sin (90^ +' A) = cos A, and cos (90** + A)
= - sin A, for any value of A from 0" to 180^
A —
7. Assuming generally that 2 cos' -x- = 1 + cos A, and
2sin^ -^ = 1 - cos A, showthat ^/2cos „ = - a/1 + cos A
and v2 sin o = — Vl - cos A, when A lies between 360'
and 540".
A
8. Given cos A = 1 *• 2 sin' -^, show that sin A
« . A A
= 2 sin -„ cos-x .
J^
9. Hence show that 2 cos ^ = - n/1 + sin A
j^
— Vl - sin A, when „ lies between 135° and 225°,
Solve the following equations : —
10. Cos' A + f cos A = tV
11. Tan + 5cot0 = 6.
'2
12. Sin A + sec A = -j-^ + ^.
13. 2 cos' A = 3 sin A.
14. Sin (A + B) = cos (A - B) = ^1-
15. Tan' A = 2 sin' A.
16. Sin (3 A + 75*') = cos (2 A - 15°).
' ' 5
17. Sec + cos = 5~7^. tan O.
18. Tan + cot = 4.
5 z
3M
CHAPTER T.
I.OGAKITHXS.
SOl DfF. — Tk logKiUkm cf & nnmlwr to a girm Ink
is the index cf the poorer to wbidi the 1»tt niHt be nivd to
oblftiii the nmnber.
Thai, wenuijobtein ihe mmiben 1, 10, 100, 1,000, 10,000,
kCf hj namg the beae 10 to the poven 0, 1, 3, 3, 4, fe,
' i«^»ectiT6ij ; and henee^ bj the aboTe definition, ve haT»—
Logl = 0,kgl0 = 1, kg 100 = 2, kg 1,000 = 3,^
!• It !• It
the suffix 10 being added to the word /^ to indicate that tlie
base is 10. It is nsoal, howerer, in oommon kgarithms to
omit this suffix ; and hence, when there is no base eaqveaed,
the student will understand 10.
Again, the numbers 1, 2, 4, 8, 16, kc^ maj be obtained
by finding the values of 2*, 2', 2^, 2*, 2*, ^, rei^iecti?elj,
and hence we hare by definition —
Log 1 = 0, log 2 = 1, log 4 = 2, log 8 = 3, &a
2 2 2 2
Bo also we find- log 16 = 2, log 125 = 3, log 81 = 4, ix.
4 5 S
Ex. Find log 256, log 216, and the logarithm of 9 to base
4 36
Log 256 = log 4* = 4, by definition.
4 4
Log216 = log6» = log(6«)f = log 36^ = J, by definition.
86 86 36 36
Log 9 = log 3« = log (V3)< = 4, by definition
Characteristics of Ordinary Logaritlims.
21. Def. — ^The characteristic of a logarithm is the integral
part of the logarithm, and the fractional part (generaUy ex-
pressed as a decimsSi^ \a es^^ t\ie msAtissa.
CHABACTEBISTICS OF ORPINABT LOGABITHHS. 355
1, Nwmhera containing integer digits,
Eveiy number containing n digits in its integral part must
lie between 10"''* and 10".
Thns, 6 lies between 10® and lOS 29 lies between 10^ and 10',
839 lies between 10' and lOS &c,
Henoe the ordinary logarithms of all numbers having n
integer digits lies between (n - 1) and n.
The integral portion or characteristic of the logarithm of a
number having n integer di^ts is therefore (n - 1).
Hence we have the following rule : —
Rule 1. — The characteristic of the logarithm of a number
Itamng integer digits is one less tluin the number of integer
digits.
Thus, the characteristics of the logarithms of 32, 713*54,
8-7168, 56452, 73607*9 are respectively 1, 2, 0, 4, 4.
2. Nwmhert less than unity expressed <m decimals.
All such numbers having n zeros immediately after the
decimal point lie between ^and ^, or between lO*"*
and 10-«« + ''.
Thus, *3 lies between 1 and *1, or between 1 and j^t or 10® and
10-^;
'027 lies between *l and '01, or between —• and r^j, or 10"*
andl0-«;
'000354 lies between '001 and "0001, or between -L And -i-,
10* 10*
or 10 - * and 10 "" *, and so on.
•
Hence, by Def., Art, 20, the logarithm of any number hav-
ing n zeros immediately after the decimal point lies between
— n and - (n + 1). Hence, the logarithm is negative, and
the integral part of this negative quantity isn. It is how-
ever usual to write all the mantissae of logarithms as positive
quantities, and the negative integral part of the logarithm
will be the next higher negative integer, viz., - (w + 1),
We have therefore the foUowing rule :—
Bulk 2« — The characteristic of the logarithm qfa numb^*
less than unity, and expressed as a decimoX) \» IKa 'aa^^q^V'c^^
356 PLANE TRIGONOMETRY.
integer next greater thorn the number of zeros immediatdy
aft&r the decimal point.
Thus, the characteristics of the logarithms of '3, '0076,
•02535, -7687, are respectively -1,-3,-2,-1.
22. The loga/rithm of the product of two numbers is the
SUM of the logarithms of the numbers.
Let 971 and n be the numbers, and let a be the base.
Since m and n must be each some power of a, integral or
fractional, positive or negative, assume —
Z V ( Theiij ^7 definition of a logarithm,
m =
n
X = log wi, and y = log n.
Now we have mn = a*,a* = a*+ ", and hence, by definition,
log WW = 05 + y; we therefore have —
\ log (mn) =« log m + log^n, Q.E,D.
Cor. This proposition may be extended to any number of
factors.
Tbus, log^ (mnpq) = log^ m + log^ n + log^ p + log^ g.
23. The logarithm of the quotient of turn numbers is found
hy subtracting the logarithm of the d&fumwMUor from the
logarithm of the numerator.
Assuming, as in the last Art., we have —
X = log^ m,y = log^ n.
Also, —=— = «*-»', and hence, by definition,
771
log — = X " y; yre therefore have —
°a n
m
24. The logarithm of the power of a number isfcyumd by
jfULTiPLYXNG tM logarvt^m of tM mmber hy the index of ihfi
power.
CHARACTERISTICS OF ORDINARY LOGARITHMS. 357
Let it be required to find log^ N^.
Assume N = a*, and therefore x = log^ N.
We have N'' = (a*)^ = o'* and hence, by definition,
log^ Np =: px = p log^ N. Q.E.D.
Ex. 1.— Given log 2 = -3010300, and log 3 = 4771213,
find the logarithms of 18, 15, -125, 675.
Log 18 = log (2 X 32) = log2 + 2 log 3 = -3010300 +
2(4771213) = 1-2552726.
Log 15 = log (3 X V*) = log 3 + log 10 - log 2 =
•4771213 + 1 - -3010300 = M760913.
Log -125 = log(^3) = log 1 - 3 log 2 = - 3 X -3010300
= - -9030900 = - 1 +J1 - -9030900) = - 1 + -0969100,
or, as usually written, = 1-0969100.
Log6-75 = logl^ = log I' = 31og3 - 21og2 = 3{-4771213)
- 2(-3010300) =: -8293039.
Ex. 2. Find the logarithm of (^'^) ^ (;^75)^ j^^.
(2-43)« X (-032)-^
given log 2 and log 3.
We have —
Log N = I log 2-4 + 4 log -375 - 5 log 2-43 ^ .
- (-i) log -032.
1, 2» X 3 * , 3 K, 3^^ . 1 , 2«
= 2^"^-T0— *• ^ ^'^2-' - ^ ^^^10-^ •*• 3 ^'^rO'
:. J (3 log 2 + log 3 - log 10) + 4 (log 3 - 3 log 2)
- 5 (5 log 3 - 2 log 10) + J (5 log 2 - 3 log 10)
= (I - 12 + I) log 2 + (J + 4 - 25) log 3
+ (- ^ + 10 - 1) log 10
- - V X -3010300 - V X -4771213 + V ^ 1
= - 2-6590983 - 9-7809867 + 8-5 = - 3-9400850
= 4 + (4 - 3-9400850) = 4 ^Q^^^\T>^-
358 PLANE TRIGOKOMETBT. «...
Ex. IV.
1. Find the logarithm to base 4 of the following numbers:
16, 64, 2, -25, -0625, 8.
2. Find the value of log 32, log 25, log '729.
3. Given log 2 = -3010300, and log 3 = -4771213, find
the logarithms of 12, 36, 45, 75, -04, 3*75, -6, -674.
4. Given log 20763 = 4-3172901, what is the logarithm
of 20763, 2076-3, -020763, -00207631
5. Write down the characteristics of the common logarithms
of 29-6, -25402, -0034, 6176-003.
6. Given log 20-912 = 1-3203956, what numbers corre-
spond to the following logarithms:— 2*3203956, 6-3203956,
1-3203956,4-32039561
7. Given log 20-713 = 1-3162430, and log 20714 =
3-3162640, find log -2071457.
8. Given log 3-4937 = -5432856, and log 3-4938 =
'5432980, find the number whose logarithm is 3-5432930.
9. Given log 1-05 = -0211893, log 2-7 = 1-4313638, log
135 = 2-1303338, find the value of log 1?^^~-5|-1^.
10. Given log 18 = 1-2552725, and log 2-4 = -3802112,
find the value of log -00135.
11. What are the characteristics of log 1167, and log 1965?
12. Having log 2 « '3010300, and log 3 = '4771213, find
X when 18* = 125.
CHAPTER VI.
\ THJB USE OF TABLES.
25. Tables bave \ieexi iottaftd of the logarithms of all num-
bers from 1 to 100,000, «ltA^^ ^«J\tiq^ ^^-^ Ww they are
THE USS OF TABLES. ^9
practicaUy used. We shall oot enter here upon the method
of forming the tables themselveB.
The fi^owing in a q)eciineTi of the vnj in -which the
logarithmB of numliers are usually tabulated: —
No.
!
3
.
6
A
7
«
A
D.
TBon
9<HS46S
mW
r-m
tm
MSK
fir4ft
«7M
nxiA
KWIl
Hlfif
87W
S
tz
S348
wl
a
Bil
ss
ss
SS74
vm
ft48S
OffllT
*mn
Ai«)
W4S
sooo
0009
1117
1171
1228
ia34
1383
—
ri{M >:{
B
il
la
22
57
22
^
«
"
ThuH, if the nomber cooaist of four figurea only, we have
simply to copy out the figures in the column headed 0, prefix
a decimal point, and the proper charaoteriafic.
Ex. Log 7991 - 3-9026011, log 7-995 = -9028185.
When we speak of a number consisting of four figures only,
we include such numbers as -003654, -07682, Ac., the num-
ber of zeros immediately fallowing the decimal points not
being counted.
Thus, log -07997 = 2-9029271
log -007992 = 3-9026555.
When the number contains &ve figures, as, for instance,
79936, we look along the line containing the first four figures
— viz,, 7993 — of the number until the eye rests upon the
column headed 6| the fifth figure. We Uien take the first
three figures of the cdunin headed 0, and affix the four
figures of the column headed 6 In the horizontal line of the
first four figures of tiie number.
Thus, log 79936 = 4-9027424
log -079927 = 2-9026936.
It will be seen from the portion of the Ic^ritimic table
above extracted, that when the first three figures of the
Icffaritbia— viz., 902 — have beeuoiKie -i^niiXaiL^^Oas:^ vc%-u^
360 PLANE TRIGONOMETBT.
repeated, but must be understood to belong to every four
figures in each column, until they are superseded by higher
figures, as 903. When, however, this change is intend^ to
be made at any place not at the commencement of a
horizontal row, the first of the four figures corresponding to
the change is usually printed either in different type, or, as
above, with a bar over it. Thus we have above 0031,
indicating that from this point we must prefix 903 instead
of 902.
Thus, log 79-986 = 1-9030140,
log -0079987 = 3-9030194.
26. To find, Hie logarithm of a number not contained in Hie
tables,
Ex. Find the logarithm of 799-1635.
Since* the ma/ntissa of the number 79916*35 is the same as
the mantissa of the given number, and that the first five figures
are contained in the tables, we may proceed as follows —
(1.) Take out from the tables the mantissa corresponding
to the number 79916. This is -9026337.
(2.) Take out the mantissa of the next higher number in the
tables— viz., 79917. This is -9026392.
(3.) Find the difference between these mantissse. This is
called the tabular difference, being the difference of the
mantiss8B for a difference of unitt/ in the numbers. "We find
tab. diff = -0000055, which we caU D.
(4.) Then assuming that small differences in numbers are
proportional to the differences of the corresponding logarithms,
we find the difference for -35 = -35 x -0000055 = -0000019,
retaining only 7 figures. This is often called d,
(5.) Now adding this value of c^ to the mantissa for the
number 79916, we get the mantissa corresponding to the
number 79916-35.
(6.) Lastly, prefix to this mantissa the proper charac-
teristic.
The whole operation may stand thus —
M. of log 79916 = -9026337.* (1).
M. of log 79917 = -9026392
Tabular dtfiference or D" = -0000055
^ ♦ Thus log 79916-S5 tr \o% VV^ ^n^\^•^^ ^'i. ^ v^'Wa635..
THE USE OF TABLES. 361
HenCe, difference for '35 or d
= '•35 X -0000055 =-0000019 (2).
Hence, adding (1.) and (2.) —
M. of log 79916-35 = -9026356.
.-. log 799-1635 = 2-9026356.
or better thus, omitting the useless ciphers —
M. of log 79916 = -9026337
M. of log 79917 = -9026392
.-. D = b5
Hence, <^ = -35 x 55 = 19
.-. M. of log 79916-35 = -9026356, as before.
In the next article we shall show how the required dif-
ference may be obtained by inspection from the tables.
27. Proportional parts.
We saw in the example just worked that the tab. diff.
(omitting the useless ciphers) is 55, and if we examine the
table in Art. 25, we shall find the difference between the
mantissse of any two consecutive numbers there to be 54 or
55 — generally 54. The nimiber 54 is therefore placed in a
separate colimm at the right of the table, and headed D.
The student wUl understand that the tab. diff. changes from time
to time, and is not always 54 or 55.
Now assuming as in (4.) of the last article, we have-
Diff. for -1 = 54 X -1 = 5 Diff. for -6 = 54 x -6 = 32
„ -2 = 54 X -2 = 11 „ -7 = 54 X -7 = 38
„ -3 = 54 X -3 = 16 „ -8 = 54 X -8 = 43
„ -4 = 54 X -4 = 22 „ -9 = 54 X -9 = 49
•5 = 54 X -5 = 27
We find therefore the numbers 5, 11, 16, 22, 27, 32, 38, 43,
49 placed in a horizontal row at the bottom marked P, in
the columns respectively headed 1, 2, 3, 4, &c.
Hence, if we require the difference for (say) -7, we take
out the number 38 from the horizontal row marked P, in-
stead of being at the trouble to find it by actual computation.
The following example will illustrate how we proceed
when we require the difference for a decimal containing more
than one de^unal ^gare. No explanaUon la n^^^^^^*^
362 PLANS TBIG0K03IETBT.
Ex. Find log 7994-3726—
M. of log 79943 = -9027804
Diff. for 7 = 381
2 = 11
32
99
6
V
/. M. of log 79943726 = -9027843
Hence, log 7994-3726 = 3-9027843.
28. Having given tlie logarithm of a rmmher to Jind th
number.
After the explanations of Art. 26, the method of working
the following examples will be easily understood : —
Ex. 1. Find the number whose logarithm is 1-9030173.
Taking from the tables the mantissae next above and below,
we have —
•9030194 = M. of log 79987
•9030140 = M. of log 79986 (1).
54 = D.
Again, -9030173 = M. of log N (2).
Hence, subtracting ( 1 ) from ( g ) —
33 = c/, the difference between the logarithms of the re-
quired number and the next lower.
33
Now ^-r = -61, the difference between the next lower
54
number and the required number.
Hence^-9030173 = M. of log 79986-61 ;
.-. 1-9030173 = log -7998661 ;
.-. '7998661 is the number required.
Ex. 2.* Find the valueof£:2?3)!_MlOW23)_*
^^, . J.L1X uxxtjvmucwx (1-32756)*
We have —
log N = 31og 1023 + ilog ^00123 - 4 log 1-32756.
Now, 3 log 1-023 = 3 X -0098756 = -0276268
i log -00123 = i (3-0899051)
= i (4 + 1-0899051) ^ 1-2724763
* The logaTit\mi% 'CL^^Vsi^»\l^ ^^i&siCK^^^ ^3:&^A^^\iL^s5so^t\lfi tables.
TRIGOKOMETEICAL ^TABLES. 363
.-. adding, 3 log. 1-023 + J log -00123 =T;-3001031
Again, M. of log 13275 = -1230345
and diff. for -6 196
.-. M. of log 13275-6 = -1230641
.-. 4 log 1-32756 = 4 X -1230541 ^_ -4922164
Then, subtracting, log N = 2 5078867
Hence we have, -5078867 = M. of log N,
and -5078828 = M. of log 32202 ;
39 = d,
also 135 = B,
39
; ^^^ 135 = '^^•
.-. -5078867 = M. of log 32202-29 ;
.-. 2-5078867 = log -03220229.
Hence -03220229 is the number required.
Trigonometrical Tables.
29. We use trigonometrical tables much in the same way
as we do tables of ordinary logarithms of numbers.
Tables have been formed of natural sines, cosines, «fec., and
also of logarithmic sines, cosines, k/Q. It is with the latter
only we shall now deal, though many of our remarks apply
equally to the former.
As the values of the natural sines and cosines of all
angles between 0° and 90° are (Art. 11) less than unity ^ it
follows (Art. 21) that their logarithms are negative. To
avoid, however, printing them in a negative* form, and for
other reasons, it is usual to add 10 to their real value, and
hence in using them we must allow for this. The same
thing is also done in the case of logarithmic tangents, co-
tangents, secants, and cosecants.
We generally express the tmie logarithmic sine by log sin,
and the lobular logarithmic sine by L sin.
Hence, we have, log sin A = L sin A - 10,
log cos A = L cos A - 10, &c.
It muse be remembered in using the tables that, although
(Art. 11) the sine^ secant, and taagent oi «Ji «xl*^<^ xitv^^^xjij^^^iss^
364
PLAKE TBIGOKOMET&r.
the angle increases from 0*" to 90*, yet tHe cosine, cosecant,
and cotangent diminish as the angle i/ncrecues.
Hence, when any angle is not exactly contained in the
tables, we must add the difference in the case of a sinet
secant, or ta/ngent; but subtract it in the case of a cosine^
cosecant, or cotamgent.
And, conversely, when the given logarithm is not con-
tained exactly in the tables, we must in the case of the sim^
secamt, or tamgent take out the next lower tabular logarithm
as corresponding to the angle next lower; but in the case of a
cosine, cosecant, or cotangent, we must take out the next
higher tabular logarithm as corresponding to the angle next
lower in the tables.
We shall a^svmie that small differences in the angles are
proportional to the corresponding differences of the loganthmic
trigonometrical ratios
Ex. 1. Find L sin 66^ 28' 2^.
Referring to tables, we have —
Lsin66^28'
Tab. diff. for 60^ or D = 836
.-. diff for 24' or (f = ^ x
60
.-. Lsin66''28'24'
Ex. 2. Find L cos 29° 31' 28''.
Now L cos 29" 31'
Tab. diff for 60' or D = - 716
= 9-9209393
836 = 334
= 9-9209727
= 9-9396263
diff for 28'
= -|l.m = L
334
.-. Lcos 29° 31' 28" = 9-9395919
Ex. 3. Find the angle A, when L tan A = 9-8658585
We have 9-8658585 = L tan A.
Next hwer, 9-8657702 = L tan 36° 17',
Also,
d
883 = difference or d,
2648 = tab. diff for 60' = D,
And - X 60" =
D
883
2648
X 60^ = 20".
Hence, ^-^^^%^%^ ^ "l^Xasv^'S \Y ^^'.
t
TBIGOKOMETBICAL TABLES. 365
Ex. 4. Find the angle A, when L cot A = 10*0397936.
We have, 10-0397936 = Loot A,
Next higher, 10-0399770 = L cot 42' 22',
1834 = difference or dj
Also, 2537 = tab. diff. for 60"' or D,
And4 X 60^ = I||i X 60^ = 43^
Hence, 10-0397936 = Lcot 42° 22/ 43^.
Ex. V.
1. Given log 47582 = 4-6774427, and log 47583 =
4-6774518, find log 47-58275.
2. Given log 5-2404 = -7193644, and log 524*05 =
2-7193727, find log -5240463.
3. Given log -5614 5 = 1-7493 111, and log 56-146 =
1-7493188, find log Ay-05614581.
4. Given log 61683 = 4-7901655, and log^ 616-84 =
2-7901725, find the number whose logarithm is 2-7901693.
6. Find the value of (1-05)^, having given log 1-05 =
-0211893, log 20789 = 4-3178336, and log 20790 = 4-3178545.
6. Find the compound interest of XI 20 for 10 years at %
per cent, per annum, having given log 1*04 = -0170333, log
14802 = 4-1703204, and log 14803 = 4-1703497.
7. A corporation borrows X8,630 at 4| per cent compound
interest, what annual payment will clear off the debt in 20
years?
Log 1-045 = -0191163, log 4-1464 = -6176712, and
log 4-1465 = -6176817. ,
8. Fmd the value of -i (44:72ldY ' ^^^^S given
Log 1-032 = -0136797, log 34722 = 4-5406047.
Log 3762 = 3-5754188, log 26202 = -4183344.
Log 34721 = 4*5405922, logi&^Q^ «^ ^«^\^'^Vl^.
366 PLANE TRIGONOMETRY.
9. Find Lsin 32" 28' 31", having given Lsin 32" 28' =
9-7298197, and Lsin 32" 29' = 9-7300182.
10. Find Lcoseo 43' 48' 16", having given L sin 43" 48' =
9-8401959, and Lsin 43'' 49' = 9-8403276.
11. Required the angle whose logarithmic cotangent is
10-1322449, having given L cot 36" 25/ = 10-1321127, L cot
36" 26' = 10-1318483.
12. Construct a table of proportional parts, haying given
163 as the tabular difference.
13. In what time will a sum of money double itself at 5 per
cent, per annum, compound interest?
14. Find x when 1*03* = 1-2143, having given that log
1-03 = -0128372, and log 12143 = 4-0843260.
15. Solve the equation 2**""^ — 40 = 9-2*, having given
log 2 = -3010300.
16. Given L cos 32" 45' = 9-9341986, D = 762, find Loos
32^* 45' 12^ and Lsec 32" 45' 20'/.
17. Given Ltan 28" 38/ = 10-2628291, D = 3003, find
Ltan 28' 37/ 15^ and Lcot 28" 38/ 42^
18. Find the angle whose logarithmic cosine is 9*9590635,
having given —
Lcob24"29' = 9*9590805,
Lco8 24"3l/ = 9-9589653.
I I
I
CHAPTER Vn.
/
PROPERTIES OF TRIANGLES.
80. The sines o/tlte angles of a triangle are prcpariicnal
to the opposite sides.
We shall deagna\A \!bA cidoa opposite to ihe angles A, B,
C, I>7 the small leitene^ a, 1>> c>t««^^>sln^i.
PROPERTIES OP TRIANGLES. 3G7
Draw AD perpendicular to BC, or to BC produced.
Then sin B =
AD AD
AB "" c
And sin C =
AD AD
AC " 6 '
XT^ /I I . /o\ sinB AD AD b
Hence, (1) -f (2),-^-- = ^ = _,
sinC c h c
or
sin B sin C
(1).
(2).
It follows, therefore, from the symmetrical nature of this
sin A sinB sin C nwn
equation, that —z — = ~t — = — • V-^-^«
a
31. In any triangle^ cos C =
2ab
Taking the figures of the last article, we have —
(1.) When C 18 an acute angle —
By Euc. n., 13, AB^ = BC^ + AC* - 2BC.CD.
* CD
Now Yn = cos C, or CD = AC cos C,
Hence we have, AB» = BC* + AC - 2 BC. AC cos C,
OP c* = a' + 6* - 2 a6 cos C.
... cos C = 5l±^\
3G8 PLANE TRIGONOMETRY.
(2.) When is cm obtuse angle, as in the second figure-
By Euc. II., 12, AB'* = BC^ + AC2 + 2BC.CD;
and CD = ACcos ACD = AC cos (180^- C) = - ACeosC-
Hence, AB^ - BC + AC + 2BC(- ACcos C);
or, c^ = a* + 6* - 2 ah cos C.
/. cos C = -—^ , as before.
^ 2ab '
From iheform of this result we have also —
6^ + c2 - o*
Cos A =
CosB =
2 he '
g^ + c^ - h^
2ae
82. To express the sine of any angle of a triangle in term
of the sides.
We have —
Sin?A = 1 - cos^ A = 1 - /^Jt-^JL^Y
^ ( 2 bey - (6' + c^ r- g*)'
(2 bey
^ |(6 + cy - a^}{a^ - (6 - c)^}
(2Jcy
{a + b + c) {b + c — a) {a + c - b) {a + h -c) ,
(2 bey
Hence, taking the square root, and taking the positive sign,
because (Art. 11) the sin A is always positive when A is fiie
angle of a triangle, we have —
Sin A = ^-r-- J(a + 6 + c) (6 + c - a) (a + c - ft) (a + 6-c).
Let a + 6 + c = 2», then 6 + c - a = 2 (« - a),
a + c - ft = 2 (s -r- ft), a + 6 - c = 2 (« - c).
Hence, sin A = -^-r- J2s,2{s - a).2(« -ft).2(« - e)
^ "be *^*^* " ^^^* " ^^^ '■-^^'
FROPERTIES OF . TRIAKGL£S. 36^
From the form of this result we have also-^'
2 '. .
SiiiB=*— V« {9 - a) {« - 6) (« - c),
2 '
Bin C = — V« (« - a) (s - 6) (« - c).
88. To find the a/rea of a triangle.
Using the figures of Art. 30, we have—
A ABC = JBO. AD, or, since AD = AC sinC,
= iBC.ACsinC
= 1 06 sin C (1).
From ihe/orm of this result we have also —
A ABC = iocsinB
and A ABC = |.5csinA
The results in (1), (2), (3) express the area of a triangle in
terms of two sides, and the included angle.
We will now express the area in terms of the three sides.
We have —
A ABC = I a6 sin C, or, by last Art,
2 ,
= J a5.^ ^/8(8 - a) {8 - 6) (s - c)
= \/«(s - a) (a - 6)(« - c) (4).
84. To express the sine, cosine, and tangent of luilf an
angle of a triamgle in terms of the sides.
We have, Art. 18,
1 - 2 sin ^ = cos A = ^^
„ . ,A , b-' + c'-a' _ «» - (6 - cY
.-. -sm2 = i 2bi ~ 26c
{a + c - b)(a + b - c) _ 2 (« - 6). 2 (a - c).
2 6c ~ 2 6c '
. ,A (s - 6) (/» - c)
or, sm= 2" 6c
. A /(* - 6) (8 - c)
•••sin 2"= ^/ ^ OV
870 PLANB TEIGONOMETRY. /
Ike form of iliis result gives us also —
Again, Art. 18, we lave—
2cor^; — 1 =f cos A = ^-r- }
or, 2cps 2 = 1 + 21^ = Jbi
(a + b + c)(b + c - a) ^ 2g.2(g ~ a) ,
2 6c " 2 6c '
A 8(8-0)
or,cos»^ = — g^— .
A /s (8 - a) .^-
The ybrw of this result gives us also —
B PiT^ /£EE1
Now, (1) + (2), we have —
Sin:^
2 - -or
Cos^
2
_ lis - b ) (s - c) fs§ZA
tan^ = J^ZMZ3 (3).
4 'V 8(8 - a)
The/orm of this result gives us also —
tan 5= /£E3ZE3,tan5= /gZ^EZS.
2 V «(« - 6) ' 2 V «(« - c)
35. /n a/ny tricmgh ABO, a = 6 cos + c cos B,
Using the figures of Art 30, we have —
(1.) "When C is acute —
BD = AB cos B, and DC = AC cos C.
.% BD -V DC = AC cos C + AB cos B, ov
a = h coi^ ^ -v <i <5as^^>
PKOPBBTIES OF TRIANGLES. 371
(2.) When C is obtus(
BD = AB cos B,
and DC = AC cos ACD = AC cos (ISO** - C) = - AC cos C.
.-. BD - DC = AC cos C + AB cos B,
or a = 5 cos C + c cos B, as before.
The/orm of this result gives us also—
6 = acosC + ccosA,c = a cos B + 6 cos A.
CoR. 1. Hence sin (B + C) = sin B cos C + cosB sin C.
6 c
For we have, 1 = — cos C + — cos B, or, by Art. 30,,
a a
sin B ^ ^ sin C -r.
= —, — r cos C + —, — r- cos B,
smA sinA
or, sin A = sin B cos C + sin C cos B.
But sin A = sin (180" - A) = sin (B + C).
••. Sin (B + C) = sin B cos C + cos B sin C.
CoR 2. Hence also —
sin (B - C) = sin B cos C - cos B sin C.
For, since this result has been proved for any angles of a
triangle, it will be also true for a triangle which contains an
angle stt^oplementa/ry to B; that is, it will be true, if we put
180" - B for B. We then have-
Sin (180' - B + C) = sin (180' - B) cos C
+ cos (180** - B) sin C.
Butsin(180'-B + C) = sm (180* - B^C) = sin(B-C),
sin (ISO'* - B) = sin B, cos (180" - B) = - cos B.
Hence, sin (B - C) = sin B cos C - cos B cos C.
Note. — ^The results of Cor. 1 and Cor. 2 have been proved only
for angles less than two right angles. We shall see in Vol. II. they
are generally true.
36. In amy triangle ABC —
TanJ(B-C)=j-^%ot|A
We have. Art. 30,
SinB h si nB-sinC 6--c
S72 PLANE TBIOONO^fiTAY.
TB + B - C)
Now, sin B = sin j — o — + — 2 — I ' ^^' ^^ ^^* •'>'^^'^>
B + C B-C B + C.B-0 ,,,
= sin — 2 — ^^ — 2 — "*" ^^^ — 2 — ^"^ — 2 — ( )*
/B + C B - C)
andsinC = sin | -^ 3" / ' ^^' ^^ ^^^- ^^ ^^5,
..B + C B-C ^ B + C.B-C ,„v
= sin _^_ COB— ^ cos -__8m_^_ (3).
B + C B-C
(2) - (3), then sin B - sin C = 2 cos — s — ^^ o ">
B + C B-C
and (2) + (3), then sinB + sinC = 2sin — s— cos — s— •
^ B + C.B-C
• T> • n 2 cos — j> — Sin — o —
sin B - Bin C 2 2
•• sinB + sinC" ^ . B + C B-C
2 sin — o — cos — o —
B + C B-C / _ A\ . B-C
cot — o — tan — r^ — = cot 1 90° - ttI tan
(»• - 1)
, A B - C ,,,
= tan 2" tan — s — w'
/.v , X A. B-C 6 -c
(4) = (1), then tan 2 tan — ^ = g-^-
B-C 6-c ,, ^„^
•'. tan 2 — = T ■ cot J A. Q.KJJ.
Ex. VI.
1. If a = 5, C = 30", sin A = i, find c.
2. Find a, having given 6 = 12, c = 15, A = 60*.
3. Find tan -— , when a = 6, 6 = 7, c = 8.
4. What is the area of a triangle whose sides are 48, 52, 20?
5. Given two ^des of a triangle to be 18 and 24, and the
included ang\^ 45% ^a^ ^^ «c^»k
PROPERTIES OP TRIANGLES. 373
6. Two of the sides of a triangle are as 2 : 1, and the in-
cluded angle is 60°, find the other angles.
7. In any triangle show th at —
h cos C - c cos B = ija? - 4 6c cos B co^ C.
8. Show that the area of a triande =. \ c?. ; — i — .
° ^ sin A
9. An object is observed from two stations 100 yards
ipart, and the angles subtended by the distance between the
3bject and either station are 45' and 60° respectively. Find
bhe distance of the object from each station.
10. An observation is made from a point known to be
iistant 120 and 230 yards respectively from two trees, and
bhe angle which the trees subtend is foimd to be 120*. Find
the distance between the trees.
a sin* + 6 cos* e = m, "| -j h ^
h sin* <^ + a cos* <f> =n, > then- + -. = —
atan0 = 6tan<^, j « ^ '^
1 2* If alf hi, c] be the sides of the triangle formed by joining
the feet of the perpendiculars from the angles A, B, of the
triangle ABC upon the opposite sides, then —
?i + ii + £! = g' + 6* + gg
a^ h^ (^ 2abc '
13. A perpendicular AD is drawn from the angle A of a
triangle, meeting the opposite side BC and D ; and from D a
perpendicular is drawn to AC, meeting it in E. Show that
DE = 6 sin C cos C.
14. Show that the length of AD in the last example —
6c sin A + oc sin B + a6 sin C
^ ^ 3a '
15. Show that /J 8 {a — a) (« - 6) {« — c) = J ah, when the
triangle is right-angled at C.
16. Show that (a + 6 + c)* sin A sin B
= (sin A + sin B + sin C)* ab.
17. Show that in any triangle —
cos* A + coB*B + cos*C + 2coeA.cio^'&^q^^ -V
374
PLANE TBIGONOHETRY.
18. The sides of a triangle ABC are in arithmetical progTes-
sion; show that its area = — ^ A/(2a - 6) (3 6 - 2a).
CHAPTER VIIL
SOLUTION OF BIOHT-AKOLED TRIANGLES.
37. A triangle can always be determined when any three
elements, with the exception of the three angles, are given.
In the latter case we have only the same data as when two
angles are given, for the third can always be found by sub-
tracting the simi of the other two from two right angles
(Euc. I., 32).
Hence a right-angled triangle can always be determined
when any two elements, other than the two acute angles, are
given besides the right angle. And when one of the acute
angles is given, the other may be obtained by subtracting it
from a right angla
We have the following cases! —
Case 1, When the two sides containing the ftghi cmgU am
given.
k We shall take C as ttie right angle
in every case.
Now tan A =
a
or, L tan A -»- 10 = log a - log 6 J
or, L tan A = 10 + log a — log6...(l)*
This determines A, and we then havo
B = 90^*- A (2)*
Also, — = sin A, or log a - logc = L sin A - 10.
.*. log c =^ 10 + log a - Lsin A...
Hence the three el^me^iLXA) ^)^> <^s^^ <ktermined
,(3)i
SOLUTION OF RIQHt-ANGLED TRIANGLES. 3if 6
Case 2. TThen tlie hypothenuae and a side are given.
Let a be the given side.
We have sin A = -, or L sin A - 10 = log a — loge.
c
.'. LsinA = 10 + log a - logc (l)t
and B = 90° - A (2),
also U^ = <^ - a^ = {c + a) {c - a) ;
.-. log6 = J {log(c + a) + log(c - a)} (3).
Case 3. When an acute a/ngle and a side m*e given.
Let A, a be the given angle and side.
ThenB = 90" - A..... (1),
also — = tan B, or log 6 = L tan B - 10 + log a (2),
and — = sin A, or log a - Idg c = L sin A - 10,
or logc = 10 + log ti - L^inA.
Case 4. WJienthe hypotheniise and an acute tiitgie are given.
Let A be the given acute angle.
WehaveB = 90^ - A ;.... .;..: (1).
Also-- = sin A, or log a = log c + L sin A - 10 ...(2).
. c
And- = cos A, or log 6 = log c + Lcos A - 10... .(3).
c
It is evident, from Art 35, that when the stUgles only of a
triangle arQ known, we can determine the ratio only of the
three sides of the triangle to each other.
Ex. 1. Given A = 23' 41', a = 35, sblve the tridiigle.
This is an example of Case 3:
Wo have B = 90* - 23» 41' = 66' 19'.
Again, log b - Ltan B - 10 - log a
- Ltan66M9/- 10 - log 35
= 10-3579092 - 10 + 1-5440680
= 1-9019772 = log 79-795
.-. 3 = 79-795.
379 t^LAinS TRIOOl^OHBTBTt
Al8ol6gc = 10 + log a - Lain A
= 10 + log 35 - LBm23"4l'
= 10 + 1-5440680 - 9-6038817
=: 1-9401863 = log 87-134
.-. c = 87134.
Ex. 2. Given a = 214, h = 317, solve the triangla
. This fiedls under Case 1.
We have L tan A = 10 + log a - log h
= 10 + log 214 - log 317
= 10 + 2-3304138 - log 2-5010593
= 9-8293545,
Next lower in tables is 98292599 = L tan 34' 1';
r.d = 946.
Also, by tables, D = 2724,
K jid /jA/. 946 X 60" • oi // ^ i_
Andjg X 60^ = 2^24 "" 2V' nearly.
.-. LtanA = Ltan34'*r21",
orA = 34'r21".
Hence B = 90 - 34' 1/ 21" = 45" 58/ 39",
And similarly may be determined.
Ex. VIL
1. Given a = 32, A = 63' 45', find S.
Log 32 = 1-5051500, Loot 63' 45' = 9-6929750,
Log 15780 = 4-1981070, log 15781 =4-1901345.
2. Given c = 151, A = 37' 42/, find a.
Log 151 = 2-1789769, L sin 37' 42/ = 9-7864157,
Log 92340 = 4-9653899, tab. diff. = 47.
3. Given a = 60, c = 65, find 6, A.
Log 2 = -3010300, log 3 = -4771213,
Log 65 = 1-8129134, Lsin 67' 22/ = 9-9651953,
Lsin67'23/ = 9-9652480.
4. Given a = 73, 6 = 84, find A, c.
Log 73 = 1-8633229, L tan 40' 59/ = 9-9389079,
Log 84 = 1-9242793, Ltan 41' = 9-9391631,
Lsin 40' 59/ = 98167975, Lsin 41* = 9-8169429,
Log 111-288 = 2-0464479,
dOtUTION OF O^LIQtJlS-AKOLED TEIAKGLES. 377
6. Given B = 71' 41' 10", c = 24, find b.
Log 24 = 1-3802112, Lcos 18** 18' = 9-9774609,
Lcos 18' 19' ='^•9774191, log 2278*4 = 3-3576300,
Log 2278-5 = 3-3576490.
6. Given a = 293, c = 751, find b.
Log 1044 = 3-0187005, log 458 = 2-6608655,
Log 691-49 = 2-8397830.
7. Given a = 12, A = 30°, find b, c.
8. Given c = 10, B = 75', find a, 6.
9. Given a = 17, c = 34, find A, b.
10. Given a = 5, 6 = 5 ^S, find A.
11. Given a = 28, B = 15*", find the length of the perpen-'
dicular from on AB.
12. .CD is the perpendicular from on AB, and DE the
perpendicular from D on BC. Given B =s 60", a = 20,
findDE.
CHAPTER IX.
SOLUTION OF OBLIQUE-ANGLED TRIANGLES.
38. Gi'ven the three sides of a triangle, to find tlte remaining
parts,
"We have, Art. 31, cos A = —r , from which A
2 be
may be determined ; and from similar formulae we may find
B and C. These formulae are not however adapted to loga-
rithmic computation. We shall therefore find it generally
advisable to use the formulae of Art. 34.
A
cos vr
2 - ^l
tan ^. = fj
K"
-b){s
-c)
be
U{8
-a)
be '
'(«■
-b){s
-c)
2 'V e{8 - a)
378 PLANE TRIGOltOMET?RV.
From either of these formulse we can detennine A, and from
similar formulae determine the other a^les.
39. Given one side and two anglei^to fivd ike remaining
parts.
Of course the third ande is at once known. Let a be the
given side. ^
We have, Art. 30, b = ^^,
' ' sm A
a sinC
c = —, — V-,
sm A
both of which formula} are adapted to logarithmic com-
putation.
40. Given two sides and the included angle, to find the
remaining parts.
Let 6, c be the given sides, and A the included angle.
I " c
Wehave,Art 36,tan^(B -- C) = j ^cot^A
This formula is adapted to logarithmic coiUputation, ana
determines ^ (B - C).
We know also 1 (B + C), tot it is the cbinplethent of J A
Hence, B and C are easily determined;
,_ b sm A.
Then, Art. 30, a = — t — ^k"^ which determines Oi
sm. i>
Note. — When the two given sides are eqnal, the solution may be
efifected more easily by drawing a perpendicular from the given anigle
upon the opposite side, and so bisectmg it. By drawing a fiffure, it
is easily seen that, in this case, B = C = 90** - i A <md a = 2 {cosB.
41. Given two sides and an angle opposite to one o/them, to
find the remaining parts.
Let a, b, B be the given elements.
Then we have. Art 30, sin A = -r sin B.
fl .
(1.) Let the value of r sin B be unity.
We then have sin A = 1 = sin 90°, and .'. A « 90^
Hence the triangle is right-angled at A.
SOLUTIO:^ 05* OBLlQtTE-ANGLfiD OJttlANGLES. ^79
(2.) Let the value of t sin B be >• 1.
We then have Bin A > 1, which is impossible.
Hence, in this case, it is impossible to form a triangle
with the given elements.
(3.) Let the value of -7 sin B be •«= 1.
Then, since, Art. 15, the sine of an angle is the same as
the sine of its supplement, there are two values of A which
satisfy the equality, sin A = t sin B, and these Values are
supplementary.
Let A, A' be the two values, then the t^lation between
them is A + A' = 180''.
If a is not greater than h, then A is not greater than B,
and there is no doubt as to which value of A is to be taken.
If, however, a is greater than h — ^that is, if the given angle
is opposite to the less of the given sides, we must have A
greater than B, and both values of A may satisfy this con-
dition. This particular case, when the given angle is opposite
to ike less of the given sides, is called the amlngtums case.
"We will illustrate this geometrically.
42. The ambiguous case.
Let a, 5, B be given to construct the triangle.
Draw the line BC equal to the
given side a, and draw BA making
an angle B with BC.
Then with centre C and radius
CA equal to h describe an arc
meetii]^ BA or BA produced in.
A and A'^ and join CA and CA *
Each of the triangles ABC,
A'BC satisfies the given conditions. ^
For, in the triangle ABC, we have BC = a, AC = 6, and
Z ABC = B 5 and in the triangle A'BC> we have BC = a,
A'C = 6, and Z A'BC = B.
Again) the sides BA and BA' correspond to the two valuer
380 1PLAKB IrftlGONOMetRV.
of c which are obtained from the two values of A in tlid
.,»H.,.i.A.|.u.B(^l.«^).
And the angles AOB and A'OB correspond to the two
values of C winch would also be found.
Cor. If a perpendicular CD be drawn from C upon AA'^
and if c' and c be the lengths of £A' and BA respectivelj, it
may be easily shown that c' + c = 2 a cos £> and c' ^ c =
2 6 cos A.
Ex. vni
Solve the following triangles, having given-—
1. 6 = 12, c = 6, A = 60^
2. a = 18, 6 = 18 ^/2;A = 30^
3. a = 5 n/3^6 = 5 J2,c = f ( ^6 + J2).
4. a = 12, B = 60^ C = 15^
5. a = 3 V2 + V6,6 = 6, C = 45°.
6. a = 10 n/S; 5 = 15 n/27A = 45°.
Given —
7. 5 = 251, e = 372, A = 40° 32', find B and C.
Log 121 = 2-0827854, L cot 20° 16' = 10-4326795,
Log 623 = 2-7944880, L tan 27° 44' = 9-7207827,
Ltan27°45' = 9-7210893.
8. a = 237, b = 341, B = 28° 24', find A.
Log 237 = 2-3747483, L sin 28° 24' = 9-6772640,
Log 341 = 2-5327544, L sin 19° 18' = 9-5191904,
L sin 19° 19' = 9-5195510.
9. C = 26° 32', and a : 6 : : 3 : 5, find A, B.
Log 2 = -3010300, L cot 13° 16' = 10-6275008,
L tan 46° 40' = 10-0252805, L tan 46° 41' = 10-0255336.
10. a = 14, 6 = 16, c = 18, find A, B.
log 2 = -3010300, log 3 = -4771213,
L tan 24° 5' = 9-6602809, L tan 24° 6' = 9*6506199,
L tan 29° 12' = 9-7473194, L tan 29° 13' « 9-7476160.
HEIGHTS AND DISTANCES. 381
11. a = 3, 6 = 2 A = 60°, find B, C, and c.
Log 2 = -3010300, log 3 = 4771213,
L sin 35° 15' = 97612851, L sin 35° 16' = 9-7614638,
Log 1-3797 = -1397847, log 1-3798 = -1398161.
12. a = 5, 6 = 6, c = 7, find A.
Log 2 = -3010300, Ltan 22° 12' = 9-6107586,
Log 3 = -4771213, L tan 22° 13' = 9-6111196.
13. If c, c' be the two values of the third side in the
ambiguous case when a, 6, A are given, show that—
(c - cf + (c + c')^ tan'^ A = 4 a\
14. If a, 6, A are given, show from the equation —
6' + c* - 2 6c cos A = a';
that if c and c' be the two values of the third side —
cc' = 6' - a', and c + c' = 2 6 cos A.
15. Show also from the same equation that there is no
ambiguous case when a = 5 sin A, and that c is impossible
whena^ftsinA.
16. Having a - 6, A, B, solve the triangle.
17. Given the ratios of the sides, and the angle A, solve the
triangle.
d) A
18. If in a triangle tan | (B — C) = tan'* ^ cot o~9 show that
h cos (p ^ c.
CHAPTER X.
HEIGHTS AND DISTANCES.
43. "We shall now show how the principles of the previous
chapters are practically applied in determining heights and
distances.
We have not space here to describe the instruments by
which angles are practically measured, but we shall assume
that they can be measured to ali^ost any degree of accuracy.
382
PLANE TRIGONOMETRY.
44. To find the Imght of an accessible object.
Let AC be the object, and let
^ any distance BC from its foot be
measured.
At B let the angle of elevation
ABC be observed.
Suppose BC = a, Z ABC = 0.
Then, we have —
AC ^ ._^ AC ^
JC ^7T = tan ABC or tane.
.'. AC = a tan 0, the height required.
tlx. Let a = 200, and e = 30°.
Then AC = 200 tan 30° = 200. -t^
45. To find tlie Iieight of an inaccessible object
A. At any point B in the
horizontal plane of the base
let the angle of elevation
ABC be observed.
Measure a convenient dis-
tance BD in the straight line
CB produced, and observe
the angle of elevation ABC.
Let BD = a, Z ABC = e, Z ADB = ^.
Then, Euc. I. 32, Z BAD = e - ^.
T^T AB sin ADB AB rukP
Ssow, — - = . _ . _ or
200 J3.
3'
BD sin BAD a sin(0-0/
/. AB = a.
AC
sm <^
8in(0— <^)
(1^
Again, -jTg = sin ABC = sin e, .•. AC = ABsine,
or, from (1) AC = a^f""*
Bin (e - ^)
Ex. Let BD = 120, e = 60", ^ = 46*.
Xhen, AC s 120.??-*P! ^^5-*^.
' - sin (60" - 46°)
CTIOHTS AND DISTANCES.
383
, ^^ gin 60° sin 45^ , ^^
- ^^Q .iu 15- =120.
kA3
120, - ,^" = 60 (3 4- V3).
V3 - 1
46. To find tlie height of am. tnaccessibh object wJieii it is not
convenient to measure any distance in a iine vnth tlie base of
the object.
Let a distance BD be
measured in any direction in
the same horizontal plane as
BO, and let the angles ABC,
ABD, ADB be observed. ^'
LetBD = a,ZABO = «
ZABD = /3,ZADB = y.
Then, Euc. I, 32, Z BAD
= 108° - (i3 + y).
T,T. AB sin ADB
sin y
BD sin BAD sin {180'' - (^ + y)}
^ = ^.i^y_,. ,. AB = a._«^^ y
a
.,or
AC
sin (^ + y)
-. (1).
sin (^ + y)
AC = AB.sin o,
Again, ^!^ = sin ABC = sin a,
AB
or from (1), AC = a.!^£ «i^y D
sm (^ + y)
47. To find tJie distance of an
chject by observation from the top
of a tower whose' height is
known.
Let B be the object in the same B
horizontal plane with c the foot
of the tower, and let the angle of dejyression DAB be
observed.
Let AC = A, Z DAB = e.
381: PLANP TRIGONOMETBY,
Then we have, - -^ = cot ABC = cot DAB, or—— == oot 0)
AC h
:, BC = h cot 0, the distance required.
Cor. Suppose B to be not in the same level with C. Let
pi be the height of B above C, then B is on the same level
with a point which iah - m from A. . •. BC = (A - m) cot &
Ex. IX.
1. Find the height of a tower 200 yards distant when it
subtends an angle of 15^
2. From the top of a tower, the angle of depression of a
point in the horizontal plane at the foot of the tower was 30**.
Given the height to be 60 ft., find the distance of the point
3. The angle of depression of two consecutive milestones
in a direct line with the summit of a hill were observed to be
60*' and 30^ Find the height of the hilL
4. An object is observed from a ship to be due E. After
sailing due S. for six miles it is observed to be N.E. Find
the distance from the last position of the ship.
5. There is an object A, and two stations B and C are
taken in the same plane. Given that BC = 50, Z ABC =
60°, Z ACB = 30^ find AB, AC.
6. The elevation of a tower is found to be 45°, and on ap-
proaching 60 feet nearer the elevation is 75**. Find the height
of the tower.
7. Wishing to know the breadth of a river I observed an
object on the opposite bank, and, having walked along the side
of the river a distance of 100 yards, found the angle subtended
by the object and my first station to be 30^ Find the
breadth of the river.
8. A person, standing exactly opposite to the centre of an
oblong which measures 16 ft. by 12 ft., and such that the line
drawn from the centre to his eye is at right angles to the
oblong, observes that the diagonal subtends an angle of
60°. Find his distance.
9. The angles of elevation of the summit of one tower,
whose height is h, are observed from th^ b^e and [^ummit of
HEIGHTS OF DISTANCES. 385
another, and found to be and <t> respectively. Show that
the height of the second tower —
7 tano
= h . 7 7 — —
tano - tan^
10. From the top of a tower the angles of depression of two
objects in a direct line, and whose distance from each other
is a, are o, fi respectively. Show that the height of the
tower —
a
cot ^ — cot a
11. A person, having walked a distance a from one comer
along a side of an oblong, observes that the side immediately
behind him subtends an angle a, and the side in front an
angle A Show that the dimensions of the oblong are—
a tan a, a {I + tan a cot j3).
12. Three points A, B, C form a triangle whose sides are a,
b, c respectively, and a person standing at a point S, such that
SA is at right angles to BC, observes that the side AC sub-
tends an angle 0. Show that the distance of S from B —
= -~^ s/{a^ + c^ - b')^ + {a^ + 6^ ~ (r')^cot« 0.
13. A person walks a yards from A to E along AB the side
of a triangle ABC, and observes that the angle AEC = o ;
he also walks b yards from B to F along BA, and observes
Z CFB = p. Having given AB = c, find BC and AC.
14. A tower is observed from three stations A, B, C, in a
straight line not meeting the tower, to subtend angles a, fi, y
respectively. Show that if AB = a, BC = b, the height of
the tower —
'J.
ab(a + b)
a cot* y — (a + b) cot' /3 + 6 cot* a
When are the conditions impossible?
15. From two stations whose distance apart is a, and which
are due W. and due S. respectively of one end of a wall, the
angles subtended by the wall are each a. Show that the
length of the wall is a sin a.
5 3b
386 PLANE TRIGONOMETRY.
16. The angles of elevation of the top of a tower, whose
height is h and standing on a hill, are «, p, when observed
from two stations a miles distant, and in a direct line up
the hill. Show that if e be the slope of the hill —
a sin a sin /3
cos = y- • -; — 7 v •
h sm g3 - o)
17. The elevation of a tower was observed to be «, but
on walking in the horizontal plane a distance a at right
angles to the line joining the first position and the foot of
the tower, the elevation was /?. Show that the height of the
tower was —
a
Voot^ /3 - cot
2 •
18. The angles of depression of two objects in the same
horizontal plane, as seen from the top of a tower, are 6 and t\>
respectively, and the angle they subtend is a. Show that if
h be the height of the tower, the distance between the
objects —
= h \/cosec* + cosec^ <^ - 2 cosec 6 cosec <^ cos a. ,
ANSWEBS.
I. — Page 15.
1. 45-23, 290, -2367, -7, &c.
2. -0005, IMl, -040020, -45, &c.
3. Three thousand four hundred and sixty-seven thou*
sandths; thirty-four, and sixty-seven hundredths; three
thousand four hundred and sixty-seven millionths; three,
and four hundred and sixty-seven thousandths.
4. 35-90846, 29130-19391, 60-0239.
6. -7237, 3-32091.
6. 69-5289, 5-06679, -41481.
7. -026, 7708-71.
8. -09, 24-356706, -003627, -289, -0096, -00016384.
9. 200, -00125, 4000, -2295, -006, -5002, &c.
10. 170000. 11. 4-97 &c., 1. 12. -2.
II.— Page 23.
1 ?_? i.l i_®P_I flOT 1010 5L*®,
■*" 7 ' 9 ' 1 ' 17* see' 6 4 *
2. -4-*, W. W, W. W, W.
3. 6, 7, 12, 11, 9, 14; V» V, V. W, V/, W-
4. 7|, Ij, 7|, 16|, 71}, 31j, &c. 6. 52.
6. ih tV, ii, If, 17a. H* ; «, &C.
'?■ I, Vt, I* ; ih ^s, n ; A, ift^. «•
a /„ 6, 1, 1, 6. 9. \, il, 9i, tIt. i» !•
10. llf 11. 3. 12. 28.
388 AN8WEB8.
ni.— Page 28.
1. 4 X 11, 2», 2 X 3 X 5', 2» X 3« X 7, 3» x 47,
2 X 3 X 7 X 11.
2. 2 X 7 X 3» X 11, 2« X 5 X 43, 2» x 3', 5* x 7 x 11',
2» X 3 X 11', 2^ X 3« X 11.
6- 1, *. h ih Th, i; ih IS. u, h h iJ; h h Vt.
6. 19, 7, 13, 6, 41, 729, 14, 11, 39.
"?• I, h -i»r. tV ih f ; I, *, H. i, ih ih
9. 2 days out of every 7.
10. 2:7. 18. 15, 10, 6.
rv.— Page 31.
1. 24, 1260, 2520, 1260, 5460, 33300.
2. H, §i. A, Hi »t. ih if. «; Ml. ^!?v. Uh m;
iV^. tVs. tW; tVA. VWW. WA. iWrJ 18 J, VS'ff. bV«-
3. 12. 4. H. A. ih ii' 8. 1065f.
6. tIt. 7. IJ. & i«ff. 10. H, i, ^'y.
v.— Page 34.
1; 4. 2s>5, u. ■
2. 29H.
3. 14|U.
4. i> T^is> th-
6- 2i, if, 3ig.
6. 2i|.
7- V*-
8. 5tJj.
a iitf
10. 2jVr.
IL 12if
VI.— Page 36.
12. 1^.
1. 3}, If, 3H.
2. 5vW- 3.
5.
«• tIt. TrfTT'
6. 3iv. 6.
4.
7. iSi.
8. 2tVA, IS-iVr-
1
a 77.
10. £8. 12^8.
Jl. ^362. 68. &d. 1%. ^^V
ABITHMETIC.
389
1. 2-203125.
3. 3-703059239.
6. 1.
9. -367879.
11. 3141592.
VIL— Page 41.
2- ^ff> TVffj iii h h ?•
4. 5-098809263: 6; 15^^ A'.
7. 2-718281. 8. -321750.
10. 1015873.
12. -857142.
Vni.— Page 43.
1. 13s. 4d., 7id., 2s. 8d., £1. 5s. 84d.
2. M. 7s. 6d., 6s., Is. iT^j-d., 3s. 9d.
3. £5. 17s. 4d., £8. 15s., £25. 3s. llx\d.
4. 17 cwt. 16 lbs., 16| lbs., 6f lbs., 2^^ ^^
6. 3 m. 2 f. 62f yds., 293j yds., 4po. If yd., 82Hyds.
6. 144 days, 32 days, 4 hrs. 54' 47'^.
7. 38 lbs. 7oz. 2dwt. 15|grs., 12dwt 6Agra.
8. - 245 ac. 1 r. 27 po. 12^g»^ yds.
9. 40° 3' 28i", 35'.
10. 6153 grains.
11. 7t'j; 4 cwt Iqr. 14f lbs.
12. Iday31irs. 8'24Hf|i".
IX.— Page 45.
!• tV> tV* 2. ^V, H> 8. ^tl^, ^APi'
4. 3 lbs. 7 oz. 15 dwts., 8^% lbs.
•?. A, tV 8. 1185/^, ,iij^.
9- tWbVA- 10. ^if^.
U.6
• 6
12. Tiiiy ^7 o^ 24 hrs.
X.— Page 47.
1. 7s. 6d., 19s. 7id., 16s. 3|d.
2. XI. 5s.^ 2s. 9d., £1. 68. 3d.
390 AsrswERS.
3. 12 cwt 2 qrs., 2 cwt 3 qrs. 7 lb&, 6 cwt. 1 qr. 25 lb&
4. 1 ac. 2 r. 26f pa, ISy'^ po., 1 r. 22 pa
5. 4 qtiires 4 sheets, 11 sheets, 23 quires 4| sheets.
6. 38 galls. 3 qts. 0^ pts., 1 hhd. 60 galls. H qts. 1^ pi
nearly, 1 pk. gaU. 3 qts. If pt. nearly.
7. 4000 grains, 8 oz. 6 dwts. 16 gis. 8. -698 lbs.
9. 15s. 2cL 10. 24 tons 9 cwt. 2 qrs. 8 lb&
11. 4 cwt 1 qr. 10 lbs. 12. 3 oz. 14 dwts.
XL— Page 48.
1. -625, -53125, -55625, -928125.
2. -846153, -692307, -615384. a -36803, -11805.
4. 015625, -065476190. 5. -003125, -002232142857.
6. 003472, -040293. 7. -3125, -150625.
8. 0002232142857, -00083. 9. 1-13085317460.
10. -82285714. IL
12. -0544575
Xn.— Page 54.
t 160000, 20000, 100, 270, 2-5, 1, 3-45, 5-294.
2. 46000000, 3000000, 2950000, 1500, 395, 29&
3. 2 myriag. kilog. 2 hectog. 9 dekag., 1 myriag. 8
kilog. hectog. dekag. 8 gmu b decig., 1 myriag. 2 kilog.
3 hectog. dekag. grm. 1 decig. 3 oentig., 1 hectog. 2
dekag. grm. 2 decig. 9 centig. 6 millig., 1 grm. 5 decig;
^ centig. 3 millig., 3 gim. 4 decig. 2 centig. 7 millig.
4. 160001-2, 25100, 396-45, 203550.
6. 1000, 2-96, 2900-03, 300-1^ 3765-43.
6. 10000000000, 10000000, 50000, 349800, 4600.
7. 150, 39-4, 90-2, 1860-3, 3764, -4.
8. 10, 1-234567, '372456126, 1, -000639, -293.
9. 3203, -4, 2000-OQ^, 1^-^M, 29*34, 8300, 3457*6.
10. 18300-453, 1SS0-^4:^^A^^^^'^^'^^
ARfrfiMETlC. 391
11. 1, 73-6, 24645, 2-55, 16*95.
12. 1300, 130; 713, 71*3; 1235, 123*5; 2*9, 320, 32,
1804, 180-4.
XIII.— Page 56.
1. 16287*599 m., 10738767 m., 1322*371 sq. m., 69*548396
cub. m., 39129-99 grin., 65632*02 ares, 368*93 st, 78603-982
Ht., 288-06 fr.
2. 1600*688 m., 11696*359 m., 96*18 sq. m., 5*967600029
cub. m., 5972*935 gnn., 2450*94 ares, 409*6 st, 694003*024
Ht., 2*35 fr., 98-56 fr.
3. (1) 70*245 m., 110*385 m., 130*455 m.; (2) 486082*89
m., &c.; (3) 49 sq. m., &c.; (4) 76*190000095 cub. m., &a;
(5) 11046013-965 gr.,&c.; (6) 36000*9 ar., &c. ; (7) 41629
8t., &c.; (8) 7347660*72 lit., &c.; (9) 2932*50 fr., &c.
4. 864 fr. 91-5 c. 6. 3408 fr. If c.
6. (1) 1*56 m., 1-43 m., 1*32 m.; (2) 27788 m., 2613*75
m., 2460 m.; (3) 2*0910 sq. m., «fec.; (4) 41*82 cub. m.,&c.;
(5) 188*263 gnn., «fec.; (6) 13*2 ar., &c.; (7) 10-01 st., &a;
(8) 40446*3 Ht., &c.; (9) 293*58 fr., &c.
7. 25 fr., -15010 fr., 24120 fri, '328 fr., 1*80 fr., 2857 fr.
64 c. nearly, 16*5 fr., 12 c, '01 fr.
8. 25, 80, 2400, 1440, 480000, 14400, 4, 96, 125.
9. n50fr. 10. 45. 11. 36. 12. 3877 nearly.
XIV.— Page 60.
1. 1609*314 m. 2; 57319*8975 ft. 3. 239*613 sq. ft.
4. 862784. 5. 271*7; 6. 1128 fr.
9. 67 fn 62 c. nearly. 8. 1053268765 galls.
9. £L 3s* 4d. 10. 2204 fr. 61 c.
11. 447*39. 12. 1*0392.
XV.— Page 64.
1. ^80; 2. 8s. Hid. a 344 fr. 48 J c.
4. 39' 22 J". 6. 70. ^- *1 b:.^\v^.
592 ANSWERS.
7. £A. 9s. 7d. 8. 31 days. 9. £305.
10. 3 L 34iJ|}' P.M. 11. 32' 43tV" past 2.
12* 4 months.
XVL— Page 67.
. 1. 15. 2. 30jjft.
3. 1125 miles (take 8 kilom. = 5 miles).
4. £15,000. 6. Hi days. 6. 92tV days.
7. 2s. 8d. 8. 62-5 m. 9. £154.
10. 88 horses. 11. 7^ miles. 12. 12 days.
XVIL— Page 69.
1. £70. 2. £41. 6s. 2id. 8. £39. 8s. 4id.
4. £4. 6s. 9f |d. 6. £4. 4s. 6. £21. 18s. 5d.
7- iHi?- 8. £236. lis. 8d. 9. 3ft{.
10. £385. 11. 5} months. 12. 15s. 5jtd.
XVin.— Page 72.
1. £23. 3s. 5-856d. 2. £49. 5s. 6-84d.
3. £20. 3s. 10-149d. ,4. £102. 17s. 4-941d. I
6. 10s. 9-6d. 6. £441. 2 fl. 1 c. 1-40 m.
7. £270. 12s. l-929d. 8. £200 4- (1-045)^
9. £231525. 10. £71. Is.
11. £50{(l-05)»+ (l-05)« + (1-05)} = £165. 10s. Ijd.
12. £450 T (l-04)».
XIX.— Page 75.
1. £12. 7s. 2-4d. 2. 3ifd.
3. £4. 13s. 3-38d. 4. £4. 17s. ll-66d.
6. £3. Os. 2xWTd. 6. £10. lis. 8-7d.
7. £5: 12s. 6d. 8. £42. 10s.
9. £39. 78. 6d. 10. £35.
II £2212. lOs. Vi^ S-^^^.ni^,^V^
ARlTttMETia 393
XX.— Page 77.
1. £690. 18s. 9d. 2. £304. 14s. Sx-^d
3. £5528. 10s. nearly. 4. £382. 10s. lOd.
6. £6911. 3s. Q-rVjid. 6. £126. 14s. lOd. nearly.
7. £842. 12s. 6d. 8. £7768. 5s. 3jd.
9. £863. 7s. 6d. nearly. 10. £134.
11. £281. 5s. ^ 12. 50,000 francs.
XXL— Page 80.
1. £374. lis. lOy^jd. 2. £1069. 10s. ll^d. nearly.
3. £1773. 14s. 6id. 4. £393. 7s. 8jd. neai-ly.
6. £308-8693. 6. 20.
7. 12s. lOd. 8. £522-0411.
9. £217-7937. 10. £1000 ^ (1-05)1
11. £23. lis. nearly. 12. £369-8728.
XXII.— Page 82.
1. 5s. 5d. 2. 10s. lOd.* 3. £1, 10s.
4. 6 to 5. 6. £48. 6. 4 j gallons.
7. He loses 12^ per cent 8. £72,123. 12s. 6d-
9. £t»5*j. 10. 2i. 11. £4. 5s.
12. 33J.
Miscellaneous Examples. — Page 8S.
1. 10. 2. £3. 5s. lOd. \
3. 721 lbs., 321 kilog. * 4. nA5 + nAS; !•
6. -0078125, t3^V 6c 2933-73.
7. 33-1, 3-236. 8. £2. 16s. Ojd., he lost 12j per cent
9. 12. 10. J, 6s., -04895. 11. 0, 4, 2500, l^V
12. £29. 6s. 7id., £0. 9s. 8id. 13. '£8000 stock, £7530.
14. £-002, 34-3168. 15. 83-8967, 1-9387. ,
16. £5i3, £lliJ, £12i3. 17. -2036. 18. 240.
19. 4s. 7-4d. per ounce. %^. SX(^.\^
394 AKSWEBS.
21. £723, £3. 6s. 4d. 22. £60, £40, £100.
28. 5s., £1. 17s. 6cL 24. £1. Is. 6jd, 11^.
26. £336, t^AV. 26. /^jW,,, 9-09.
27. £3375. 28. e^\. 29. -3. -. 30. 3000 days.
ALGEBRA— STAGE I.
L-t-Page 149.
1. 8, 27, 6, - 16. 2. 6, 2, 4, 0. ^ 3. 5, 11, - 7.
4. 11, 16. . 6. - 2, - 15. 6. a 1 . 7. - 5, - 7.
a - 11,8. 9. 7, - 75. 10. 72,^ 68.
11. -2,-3. 12. - 10,32.
II.— Page 154*
1. 61, 6, 3* 2. 35. S. - 513, - 65.
4. - 1224, 30. 8. 0, 2. ^6. 1, - 27. 7. 1, - 2.
8. 1691, 0. 9. 144. 10. 1521.
11. 145. 12. - -5^,
4^1029
III.— Page 156.
1. 10 a + 3 i. 2. 11 a\ 3. 12 6 + 8 c. 4. 0.
8. 8a^ -^ ah. 6. 3aJ* - a» - 15aj - 2. 7. 4a6 - 4.
8. aj' + 2/' + «' — 3 Qcyz. 9. aJ* + ai*^ + ^.
10. a« + 6« + c» + 3a*6 + 3 a6« - a'c - a<r» - ft'c - 6c'
- 2a6c.
11. aJ* + y* + aJ*-4a^^ + 4»'«-4a5y'-4^« + 4aa?
- 4y;s' + 60*2/* + eaar* +,6^»' - Vlv^yz + 12 o^
- 12a;y«2. 12. 0.,
IV.— Page 157.
1. 4a+2b + 5c. ^. - 4a; + 2y - 6«.
ALGEBRA. 395
4. 2a* - 2a2ai» + a^. 5. 4a* + d^a'U' + 4 6*.
6. -3ic»-3«'- 3 ar^« -3 a»r».
7. fiC* - oic' - 9 a^ic" - 3 a'a: - 2 a*. 8. 0.
9. -a-6-c — c? + 6+/+5r+A.
10. 2 aj* - 8 ar»y + 12 ar'2/= - 8 oj^* + 2 y*.
11. jc* + 2 ar^y» + y\
12. a' + 6* - 2 c» + 2 a5 - 2 ac - 2 6c.
v.— Page 159.
1. - aj + 8y + 2J. 2. a» - 6». 3. 13 - 6as,
4. 5 ar» - 3 a? - 7.
5. (a + 6) + (c - c^), a - (6 — c + cQ,
|a - (6 - c)} - c?.
6. -. (6a - 7 6)- (3c - 6d), -6a*+ (76- 3c + 6d),
-{6a- (76-3c)} + bd.
7. - (4ar» - 12ar'y) - {Uxf - xf),
- 4a» + (12ai»y - 12a^ + 42/»),
- {4a» - (12ar'y - 12a^}+ 43^.
a (a» - 6«) - (c» - 3a6c), a» - (6» + c« - 3a6c),
{a» - (6» + c^} + 3a6c.
9. - (a -6 + cQa:?- (a -6-3 c)ary
- (26-c-e-/)3^.
10. (a-6 + c-cQa;-(c-c? + «-/)y
-(«-/ + 5^ - ^)«-
11. (a — 6 - cQ a; - (a - 7 6 - c) y - (6 - c + cQ «.
12. (2 - a;) a« + (a; - y)a6 + (y - «)6'.
VI.— Page 161.
1. 12 a« - 06 - 6 6S 18 ar* - 9 a;y - 35yl
2. aj» - 2 ay» + 43^, 30ar» + 49 a:»y + 9 ay - 2^.
3. a* + 2 a»6^ + 6*, a* - 6*. 4. a^ - y*, a? - 3^.
6. a» + 6» + c» - 3 a6c. 6. a? - y".
7. a» + 3a'6 + 3a62 + 6».
& 5a» + 5a*- 405a - 405.
896 ANSWERS.
9. - 7ar» + ITa? - 5a; - 2, a' - a?.
10. a« + 2 aW + 3a*6* + 2 a=6« + 6«. •
11. «&/•+ (ac + 6/')aj + (6 +/) CJB* + (c» + (^ 0? + cia^.
12. a* + (jt> - a) ar* + (^r - op) x — aq.
13. a? + (a + 6 + c) ar* + (oft + ac + bc)x + abc
VII.— Page 165.
1. ar» - 2 a;^ + y\ 9 a^ - 30a6 + 25 6^ 16 c* + 8 c'cP + rf*,
9 a* - 12ar'y» + 4y*.
2. a* - 61 3. wiW - nV, 5 a^ - 186*.
4. (a + c)2 - (6 + <^)S (a + 6)^ - (c + (f)*.
6. ar* + 4a; - 5, ar* + 7a; + 10, a? + 2a; - 15,a?-25.
7. aJ* - 37a? - 24a; + 180.
8. ^Atf{4a»6^- (a»- 6' - (r»)»}.
9. 4 (a^ + 6« + c« + (P). •
■
Vni.— Page 171.
L 4rt«-a6 + 26^ a^ - 43/^.
2. 2a* -5a'6 + ^^ d?h\ 2a? + 3a;y + 2f^.
8. oaj**-" + 6aj-"y + (»;-<'* + '•* 3/**,
4. 6a; + 42f, 5a; - 3y. 6. 1 + a + a?.
6. - 3a;- 4, - 7a; + 3.
7. a;> - 3ar^2/ + 5a^ + 27 2/^ + -^-^,
a; - oy
a? - a;V - 3a^ + ISt^ - --i^.
a; + y
^, a^ + Qi?y + a?tf + xy^ + f/^yoc^ -ot?y + a?t^ - a^ + ^.
9. oa;*" + 6y. 10. a' + a'6 + 06^ + 6».
IL 6 + c, a + 6. 12. a + 6a; + ca?.
13. a* - (p - 1) a' - (p - g - 1) a« - (p - 1) a + 1.
14. The given expression is —
ALOEBR^ 397
16. - a^*^ - a?2/*, a + a""^
17. {x + yf + {x + yfz - (aj + y)s? - «?.
20. {a - c)« -. 2 (a - c) (6 - ci) + (6 - df.
IX.— Page 177.
1. (x+ 3a) (03- 3a), (42/> + 5^) (42/^ -6^,
6 (2a + 36) (2a- 36), (2a;- 32/) (4ar» + Gicy + 92/^).
2. x(x - y) {x^ + xij + jf),
(a - 6) (a + 6) (a» + 6^) (a* + ¥),
xy{x+ y) (a? - xy + y^,
2xy'z{x + 2z) {x - 2z).
3. (a" - 2 6^) (a^ + 2 b^\ (oc" + xy + y") {a^ - xy + f),
(a + 5)^ (a - 6)^ (a + 6 + c) (a + 6 - c).
4. (a + 6 + c + c?) (a + 6 — c - c^),
(a + 6 — c + i^) (a - 6 + c + c^),
(a + 6 - c) (a - 6 + c).
6. 5 {2x + 9), 3 (2a; .+ 7), (3a + 6 - c) (a + 6 - c).
6. {^ + xy^f){(x-y){7?-'if) + 4.7?f{^-xy--if))^
{t? + xy -^ f) (x" " xy + f) {x + yf {x - y)\
7. (a; - 10) (x + 7), (a; + 1) (a; + 10),
(a - 76) (a - 86), (a; - 16) (a; + 12).
8. (oa; + 7 6y) (aa; - 6 hy\ 3a(a; - 10) (a; + 2),
ac{c - 8)(c + 3).
9. (3a; -{■ 5) (2a; - 7), (2a; + 9) (4a; - 15),
3(2a; + 3) (3a; - 8), (4a; - 7) (5 a; + 6).
10. xy{ZX'{- y){x ■{- 3 i/), a; (5 x + 3 a) (4 a; + 5 6),
(mx + p) {nx + q).
11. ai» + 2ar^ + 4a; + 8, 3a;* - 6a? + 12ar» - 24a; + 48,
a;* + 3 ar^ + 9.
12. (a + 6)2 - (a + 6) (c + c^) + (c + ci)2, a - 6 + c + d.
18. a^ + pa? ^ (p? + ra + «, 2 a*. 14. - 102, 17.
X.— Page 179.
1. aV, - 27 a»6», 16 a«6V, - a?i/rf-
398 ANSWERS.
a o* + 12rf5 + 64a«6« + 108a6» + 81 6*,
16a* + 32a^b + 24a«6« + 8a6» + 5*,
a» - 6a^b + lOaW - lOiiW + 6ab^ - 5»,
27 a» - 108 a^ft + 144 a6^ - 64 5*.
8. 16w* + 32m' + 24w^ + 8w + 1,
125 a^ + 150 ar^, + 60aj + 8,
81 a* ^ 432 a«c + 216 aV - 1152 ac» + 256 c*,
- a' - 3 a«6 - 3 05^ - b\
4. i«*+2a^+3ar^ + 2aj + ],
9a« + ft* + 16c^ + ci^- 6ai + 24 oc- 6a^- 86c
+ 2 5(£ - 8 cc?,
a« +4 6^ + 0* + 4a5-2ac- 46c,
9(a* + 6" + c^ + 2a6 + 2ac + 2 6c).
5. 1 + 3a;-5a:?+3ic*-a^, (aaj + 6y)' + 3(aaj + 6y) C8J + &C.
6. 1 + 7a; + 21a« + 35aj» + 35 a^ + 21 a» + Ta/^ + af,
(l + a;)* + 4(l+aj)W+6(i +a;)V + 4(l + a;)a:« + «»,
{a + 6aj)* + 4 (a + 6a;)' «b* + <fcc.
7. aW - 8 a^6a^y + 28 a«6^a^3/» - 56 a»6Vy» + 70 a*6Vy*
- 56 aWa^j/" + 28a'6V/ - 8a6»a:2^ + 6y,
729 a:» - 243a;*y» + 27ar»2^ - 2/»,
a*- 3a^y» + 3aj»2/»-y'.
' a"^ 6aiW+ 15 aW- 20aW + 16 a*ff»- 6 aW« + 6^1
9. a» + 3a«6 + 3ay + 6». 10. {a - c)l
11 2(1 + 3ai^)*. 12. 81a*y*(a; + y).
XL— Page 193.
1. 2fljy'«», 4aV, a? + a^. 2. 2ar» - 3a»y + 5rB»2^.
3. 5^2 - 3a6 + 61 4. 1 - 2aj + 3ar» - 4arl
6. a + 6a; + ca;^ + (M. 6. aV - 3 oaj*""^ + 4 ai"~*.
7. a; + a;-^, aar^ - a-^a; 8. 3 a;^ — ^ 5 a.
9. 36, 79, 207, 289. 10. 103-2, -024, -3, -j.
11. 4-123, 1-224, -618, 3-732.
12. -0203, 4-6, a-5Q^^. \>- \^y^> 6 a^, a + 2 6.
ALGEBRA. 399
14. cc* + 3 ar* - 7. 16. a + y - c.
16. X + x-\ xy"^ + 1. 17. 18018, IMl.
18. 2-73, -3, f . } 19. -6, -2599. \
20. 4^= 1-709. 21. 7-6457, 60-2487.
22. 4-4142, 3-7320. 23. -25. 24. 0.
XII.— Page 200.
1. oj - 3. 2. ar» + 4 a; + 3. 8. a; + 3.
4. aj + a. 6. d^ ■\- ah -k- h\ 6. a: - L
7. a; - 3. 8. 3 a; - 2. 9. 12 ai^ + 5 a;.
10. 3 a + 2 5. 11. aj - 1. 12. 2 ar» - 4ai« + a: - 1.
13. 2 a + 3 6 + c. 14. aj^ + 1. 15. a; - 1.
16. 2a? - 3aj + 9. 17. 1. 18. a + 6 +c.
XIIL— Page 203.
1. 12aWy». 2. 60 aW.
3. (a - 6) (6 - c) {c - a). 4. a^x {a? - a«).
6. (a? + 1) {x + 2) (a; + 3). 6. (a;- 6) (a? + 5) (a; - 6).
7. (2a; •+ 7)(3aj+ 8) (4a; + 5).
8. 210 (ar» + 1) (ar» + 1). 9. a' (a!* + aV + a*).
10. (a; + a) (a; + 6) {x + c). IL 1 - a;*'.
12. {x + 1) (a; + 2) (a; + 3) {x - 5) (a; - 6) (a; + 5).
13. a (a - hf {a? - aft + 5^^).
14. (a; - 1) (a; + 1) (ar» + 1) (6 a? + 5aj« + 2aj - 1).
15. (a^ - 6y.
16. 2a;(ar*-l)(3a? + 3a;-l)(2a:> + 2aj-5)(2ar»-2a; + 5).
17. 3(a; - 2Y{x + 4) (5a? + 10a? + 20a; + 18).
18. a%\a + 6) (a - h). 19. 15 (3 a; - 10) (a;* - 16).
20. (a? - y2)a (a? + 2/^) («* + 3^)- 21. a? - a«.
22. (a + 6 + c + c2) (a + 6 - c — c?) (a - 6 - c + c^).
28. (a + 6 + c) (a» + 6« + (? - 3 oftc).
24. * (a + 5 + c + c^) (6 + c + (^ - a) (a + c + cZ - 6)
400 ANSWERS.
1.
2.
XIV.— Page 206.
a; - 6 a? -^ ix + 2
35 + 6' a^ + 5a3+6'
3a; + 7 2a? + 9
7x + 2' 5a? + 2a;-2'
26
3 (a -h)(a + by 4 g^ + 6^
a2-3a6 + 6^' a + 36'
* a(y + ») - / a? - y=^ * a' - 6* a^ . ^2-
/•a-6 I- a?-a?+3a;-l'
^ SnT^- ^- 2 (a; - 1) (a? + 1)^*
a ^^
(a; +!)(« + 2y (a? + 1)
g 8a; - 20 jq 8 a? + 8
{x + 1)« (a; + 3/ (a; - 1)* (a? + 1)*
11. 0. 12. 0. 18. 1. 14. a + 6 + c. 15. 1. 16. 0.
17. - li^, 18. ^: - ^ -^ ^; 19. 1. 20. 1.
{cr - arf or + ab - b^
' (a; - y) (y - «) (a; - »)' 'a?*
28. . , , \, , -. 24. 1^ - 4aj + 5a.
(a; - a) (a; + 6) (x + c) a
28. , "%-y) 89. 1. 30. L.
(a + X) {a + yy sir
XV.— Page 212.
L 3.
2. 4
3. a + 6.
4. 6.
6. 6.
6. 3.
n.'o.
& 2.
ALGEBRA. 401
9. c. 10. a + b. IL a + b + e. 12. ^LL^LLA*.
13. abc 14- - (a + 6). 15. aie. 16. l^iij:_l\
17. 2. la 8. 19. 4. 20. *^ ~ '"^ ,.
a + — <s — (»
21. ^^Ji_?&Jl1' 22.-4- 23.a + 6 + <^
a + o ooc
24. 1
a + 6 + c
XVI.— -Page 217.
1 6. 2. -^^"^ ^ a 4a
« + o - e — a
4. mi' 5- - AV & - *.
7. 1^. & 3. 9. - Ij-
10. a iL iji. 12: — 2*2— *
15. 3. 16. - 8. 17. 1. 1& ^^.
zab
19. ( V« + Jh)\ 20.-0. $SL a? + 2oL
22. ^ 2a - 2t. 24. ^yi^ g - ad) + (c/ ^ M)cd
XVIL— Page 221.
1 10. 2. 45, 25. 3. £360, £240, £120.
4. 20, 15 miles per hour. 6. 24.
6. 36a, 48a. 7. 12. 8. 36s. 9. 30a.
10. 13. IL 10. 12. 24 Iba, 40 lbs.
5 2c
402 ANSWERS.
m
13. 13. 14. 750. 15. 15s. 16. 46^.
17. 18 miles from A. 18. -^:i\hv.
19. 15 per cent 20. 28 miles. 21. 10.
22. 4 J, 4 miles per hour. 23. 12 miles.
24. 20 miles per hour in same direction.
25. 120. 26. 6s., 4s. 27. 2, 1, |, 6 hours.
28. 98 percent 29. £1,000. 30. 38^^: miles per hour.
31. 2 inches. 32. 16 vols, of hydrogen, 8 of oxygen.
gg 1 - 100c 1006 - 1 g. ah'c - a'hc'
b - c ' b - c"' ' cc'(a'b - ab')'
j^b
c — at!
35. - 40\ 36. "^'^
XVni.— Page 228. -
1. 3, 4. 2. 5, 2. 3. 3, 7. 4. 4, 5.
5. 12, 8. 6. 2, 6. 7. 1, 2, 3. 8. 1, 2, 3.
Q 4. 1 9 in ^1 ^ " ^1 acj - OiC
». *, 1, i5. lU. -y J —T ,
aOi — OiO aoi — Orfi
11. a: = ijLf^li, &c. 12. « = '^ V - ".
2 2a
13. 4, 5. 14. 3, 8. 15. 4, 8. 16. 3, 4.
17. 3, 5. 18. 8, 1. 19. X = ? ,, &c
a + (J — 6
20. 2, 3, 4, 1. 21. 3, 4, 5. 22. x = ajL^+l:^^ ^
o
23. a; = '^^'^+/-^>. &c. 80. « = y = * = 1.
XIX.— Page 231.
1. 12, 8. 2. 37. 3. 6 lbs., 5 stonea
ALGEBRA. 403
6. 6s., 3s. 7. 220 and £2. Us.
8. 200, 300, 400. 9. 48, 23, 18. 10. 5, 3, 4^.
11. 4, 6, 8 hours. 12. 6, 10, 18. , 18. 7i, 12, 3.
14. 4, 3 miles per hour. 15. 16, 32, 48 miles per hour.
16. OyC + h^y + CiZ = diy a^ + b^ + c^ = d^,
a^ + bg^/ + c^ = d^f from which x, y, z.
ALGEBRA— STAGE IL
I.— Page 300.
L ±3,
2. ±4.
3. ± 5.
4. ± \.
6. ±2.
6. ± (a + 6).
7. ± 1.
8. ± Ja" - b\
9. ± V2a6-6».
10. ± 3^^ V37
o a
^^'Jl'-
2 + 1
+ 3 6*
12. ± " V-
13. 2, 3.
14. 9, - 8.
15. 1, - f.
16. 4, - i.
17. - 1, - *.
18. - 5, - |.
19. i (6 ^/6» -
4ac).
20. 5, 3^.
2L 7, - 6.
22. 6, - f.
23. 1, "'^ - <f +
be
6)c
24. b, a.
25. a+b, "^ ,.
a + b
26. a, 6.
a — a —
e)f^ a* + cA + ¥
a - 6
29. 4, tV
a-b '
a* - a6 + 6»'
30. 5, - If.
81. 3, - 6.
82. 2, - 3j.
33. 2, - \ii.
84. 5, i.
85. 8, - f.
36. 3, - f J.
87. 2, 2i.
38. 4, - t
39. 15, - L
40. 2, 0.
\\. ^, ^ab.
404 ANSWERS.
42. a, 3| a. 43. 0, "" "^ f" + \[ ^
(o - a)b
45. a + 6 + c, =-.
a —
IL— Page 306.
1. ± 3. 2. 2, - ^V- 3. ±5, ± ~g.
4. 2, - 5. 6. ± \. 6. </2, ^ 7. 4, 69.
8. ± 4, ± ^7. 9. 3, - 4, 3 ± JW. . 10. |, 4.
11. - 2, - 2 ± V3. 12. %
13. 4, ^, - i (- 13 ± n/145). 14. 0, 5 - a.
lb. ± \a ^|T. 16. I * Jb-2asfb+ |-
^3- C^iJ- ^- * '• * ^'^^•
21. 2, - J, i(- 11 ± n/97). 22. 4, |.
23. 5, 3J, J (- 7 ± J53). 24. 3a.
25. 3, - 2. 26. 4, - 14^, i (- 19 ± V74T).
27. 3, - 21, if. 28. i (1 ± n/4 a - 3).
29. ± 1, anda!* + !B» + 1 = 0. 30. ± /?L±AjL?.
St a, i { V4c'- Sa" + 2a6 + 6* - (a - 6)}-
d)3. « = ni is oue 8o\a\.voxu
ALGEBRA. 405
83. x = a-rb + cla one solution.
34. re = ~, and a* + ax + x^ = 0.
35. X = is on« solution.
a + c
36. ic'-a; + l = gives two solutions.
III.— Page 310.
1. a; = 2, 3, y = 3, 2. 2. a; = 5, - 3, y = 3, - 5.
3. re = 3, 4, y = 4, 3: 4. a; = 2, 4, y = 4, 3.
5. a; = 5, - 2, y = 2, - 5. 6. ± 7, ± 6.
7. ± 6, ± 3. a 4, 2. 9. ± 2, ± 7.
10. ± 4, ± 3. 11. « = ± 3, -^,y = ± 2, - -^.
12. a; = 9, 4, y = 4, 9. 13. a; = 2, 8, y = 8, 2,
14. a; = 3, 2, y = 2, 3,
2 2
~ <r±^"F^^ ^ " a + n/2 6" - a»
16 +-<t=iiE- ^ <" - ^)^
■ ~ J2 («' +1)' ~ 2 Va" + 1
18. aj = 2, 3, y = 3, 2, 19. a? = 2, 3, y = 3, 2.
20. a; = 2, 3, y = 3, 2 21. a; = 5, - 3, y = 3, - 5.
22. X = ^^ n/ 2 ^ - « _!, 23, a; = 2, 3,y = 3, 2.
24. X = ± 8, qp V n/I?^ y = ± 2, ± s/h
25. a; = 4, 9,y = 9, 4. 26. a: = 4, 1, y = 8, 0.
27. a; = 4, 3, y = 3, 4. 28. ± 3, ± 1.
29. a; = 0, 3, - HH. V ^ ^.% \\\\^
406 AKSWERS.
30. 0,-1. 3L a: = 4, Viftf , y = 3, - %V-
32. a: = 5, 6, y = 4. 33. a: = 3, 1, y « 1, 0.
34. a: = 3, - W, y = 1, - }|.
36. a; =r 2, 7, y = 4, - 14.
36. X •\- y = c, and ay = fix + ay.
37. a? = 0, a — 5, y = 0, a — 6.
38. V ± 9 Jih \ ± Vlt-
39. X = y (-ZJZffiZ^ZS
^ 2 (6 + c - a)
40. 1, 2, 3. 4L ± 1, 0, 0.
42. « = y = * = O.alsol = >/l + ^ - IV
a:y \6' c" a-/
^^Ka'^-i-y'""-
43 ^ ^ <^
4^a« + 6^ + c«' 4/a' + ^ + ?* 4^a» + 6» + c»'
44. a: = y = ar = w = a + 6.
45. 7' T-, dca
(a + 6 + c) * (a + 6 + c) *
46. a: = 6, 4, y = 4, 6, « = 5.
IV.— Page 315.
1. 9, — 6. .2. Numerator 25|, denominator 3,
3. 35 or 23. 4. 10, - 16. 6. 78, - 68 ft
6. 28., 9s. 7. 20 hours. 8. 48 shillings.
v.— Page 318.
1 a\ a\ a*, a. . • 2. a + a^ (a - »)*.
ALOEBRA. 407
5. xy"'^ +05"^+ 1. 6. a^ + ah^ + 6*, a* + 6*.
7. ic - a:*y. 8. a? + 6* + c* - a^h^ - a^c^ - 6*ci.
9. a^h'^ - 6i 13. a* + 6* + c*.
14. 20;*/ - 3a;y* + 2a;*. 15. a^i" ^ ^ 2 + 2a-»6*
16, a^ + 3a;^ - r. 17. ar'y-i + 1. 18. (a^ + 6«)^
VL— Page 325.
1. re*, a* 6^, x^y^t aH*-
2. ajya "■ *6 ~ *, c^Sx "*y~*, a'mojn'
3. ^/27, VT4, n/v, -^9. 4.- >/32, ^277^32; a/v.
5. n/9^, V^, v^a' - a?. 6. ^4, .^27;
7. jys, jysr & vs, uns. 9. "Va-, ^6-.
10. ^{a + x)\ 4^(a - x)\ 11. a^,b^'
12. 2 VS; 2 >^'6, 6 a/7, 3 -5^37
13. 2a^/^TT,b'^f^im^, j^ ^^T^
1 *
14. - - a, 2! — aM.
15. -„ - ^/3^, -^ 5/(«? - aV-
oa; a: — a ^ '
16. 4 n/3, 5 4/7. 17. ?2LZ_._jL5 ^^g^^
18. (3a«t - 2a"^'- 12)4^.
19. a* - 6*, a ^ 6. 20. {x + 2/)*, a« - b^S.
408 ANSWEBS.
22. a - ft* + c* - (ft + 2 aici - 26M.
23. X - ic^y^ + y.
24. a* — at^yi + a;*y* — a!^^ + a^^f - y*.
25. J >/l5, ^ a/^, i 4/T75.
26. 3(2 - V3), }(3V2 - 2^/3), i(j5^ ^3).
27. 1(2 + ^/2 - V6), J (2 n/3 + 3 v^ - VSO), 5 + 2 V^-
28. a (x^ - a:*yi + a;4y* — ary 4. a;%^ - yJ),
3»5 (3^ - 3^ 2^ + '3?. 2^ - 3 . 2 + 3*. 2* - 2»),
29. 2 + V7, V5 + V3, 5 - V5.
30. V6 +'a/2, V5 - V3, a/15 - Vs".
YIL— Page 330.
1. a' + 6* : a' - 6' is the greater, if a s» (•
2. The former. 8. 6 - — .
c
4. (1 - y) (1 + ar) : 1 + a:*. 6. 30, 35.
6. Less than 3 when x lies between 2 and 3 ; greater than
5 for all values beyond these limits.
8. X satisfies the equation —
4 ar» - (a + 6 - 6 c)a? + 2 (c» - a5) = 0.
9. t 10.1. 14. 26- a, ^1 ""^
a^ 2 a - 6*
PLANE TRIGONOMETRY. 409
PLANE TRIGONOMETRY.
*
I.— Page 335.
1. 43-75(y, ir 50' 3-03". 2. 88^ 88^ 88-8^
3. a n the number of grades.
4. 180 - {a +1^6) degrees, 200 - (^ a + h) grades.
5. i. 7. |. 8^ 25^, 75^ 90^ 9. 576^ 648^.
10. 2nd, 1st, 4th, 3rd, 4th. 11. ^^ -^. 12. ^^^.
qa
a
IL— Page 341.
1. Cos A = -j^j, tan A = y ike.
2. Sin A = — r= — J cos A = — ,- , <fec.
n/T201' V1201'
17. |. 18. ^, (2 ^/89 - 6). 19. ^4-^'
20. 2 + J3. 21. 00 , - 4^' 22. 1.
24. Sin A = /aZK+3.
III.— Page 352.
2. From 0° to 90^ + ; from 90° to 180^ - ; from 180° to
270°, + ; from 270° to 360°, - .
3. Cos A + sin A is + , + , - , - , as A lies between - 45°
and 45°, 45° and 135°, 135"^ and 225°, 225° and 315°
respectively. The corresponding signs of cos A -
sin A are + ,-,-,+.
4. Cos 2 A is + , - , + , - respectively, as in Question 3.
'ij
410 ANSWEBS.
10. A = 60^ 11. A = 45^ and tan A = - 6.
12. A = 30^ 13. A = 30^
14. A = 45% B = 15^ 16. A = 0% 45% 135**.
16. A = 6^ 17. A = 60*=^. 18.e = 2± Ji.
IY._Page 358.
L 2, 3, J, - 1, - 2, 1-5. 2. 1, 6, 1-5.
3. 1 0791812, 1-5563025, 1-6532125, 1 8750613,
2-6020600, -5740313, 1-8239087, "2-8696761.
4. 1-3172901, 3-3172901, 2-3172901, 3-3172901.
0. 1, 1, «$, o.
6. -020912, 2091200, -20912, 20912. 7. T-3162760.
8. 3493-768. 9. -2427189. 10- 31303338.
11. 6, 5. 12. 1-67.
1. 1-6774495.
4. -0616835.
7. £663-449.
10. 9-8401608.
v.— Page 365.
2. 6-7193696.
5. 2-07892.
8. 2620-248.
11. 36^26^30^
3. r-8213310.
6. ;^l77-63 nearly.
9. 9-7299222.
VI.— Page 372.
I. c = 12-5. 2. V189L
4. 480. 6. 108N/2r
9. C = 50 j6U""^-^\^
10. a = 10 J9i9.
3. Tan:^ = ^\.
6. B = 90% C =
\^^ (^sJZ- IV
30%
PLANE TRIGONOMETRY. 411
VII.— Page 376.
I. J = 15-78065. 2. a = 92-34057.
3. 5 = 25, A = 67»22'49^
4. A = 40°59'32^c = 111-288.
5. h = 2278438. 6. 6 = 691-49.
7. 5 = 12 J3, c = 24.
8. a = iiJQ - J2), 3 = t(V6 + J2),
9. A = 30°, b = 17 J3. 10. A = 30°.
11. CD = 7(^6 - J2), 12. BE = 5^3.
VIII.— Page 380.
1. a = 6 V3, B = 90°, C = 30°.
2. B = 45° or 135°, C = 105° or 15°, c = 9 ( ^6 ± J2).
3. A = 60°, B = 45°, C = 75°.
4. A = 105°,& = 6 (3^2 - je), c = 12(2- ^3).
5. A = 75°, B = 60°, c = 2 ^6.
6. B = 60° or 120°, C = 75° or 15°, c = 5 (3± ^3).
7. B = 4r 59' 23^ C = 97^ 28' 37^
8. A = 19° 18' 11^
9. A = 30° 3' 26", B = 123° 24' 34".
10. A = 48° 11' 22", B = 58° 24' 42".
II. B = 35° 15' 52", C = 84° 44' 8", c = 137-9796.
12. A = 44° 24' 56".
IX.— Page 384.
1. 200 (2 - J3). 2. 60 J3,
3. i^3. 4. ^ja.
412
ANSWERS.
6. 6 (^3.1). 3^/6 (^3-1).^^
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'William Ck>llinB, Sons, & Ck>.'8 Educational 'Works.
COLLINS' SERIES OF FIRST-CLASS SCHOOL ATLASES,
Carefully Constructed and Engra-ved from the best and latest j^utAorities, and
. Beautifully Printed in Colours, on Superfine Cream Wove Paper,
MODKRN GKOGRAPHY— Crown Series. «. d,
MY FIRST ATLAS, consisting of 12 Maps, 9 inches by 7^ inches,
folded 8vo, in Neat Wrapper, o 6
THE PRIMARY ATLAS, consisting of 16 Maps, 9 inches by 7}
inches, 4to, Stiff Wrapper, o 6
THE POCKET ATLAS, consisting of 16 Maps, folded in 8vo, and
mounted on Guards, cloth lettered, ... ... ... ... i o
THE JUNIOR, OR YOUNG CHILD'S ATLAS, consisting of 16
Maps, 4to, with 16 pp. of Questions on the Maps, in Neat Wrapper, a o
THE STANDARD SCHOOL ATLAS, consisting of 24 Maps,
{ Crown 4to, in Neat Wrapper, I o
j THE PROGRESSIVE ATLAS, consisting of 32 Maps, 9 inches by
7^ inches, 4to, cloth lettered, 2 o
THE CROWN ATLAS, consisting of 32 Maps, on Guards, with
Index, 8vo, cloth lettered, ... % 6
THE NATIONAL ATLAS, consisting of 32 Maps, 4to, with a
Copious Index, cloth lettered, -- ...26
MODERN GEOGRAPHY— Imperial Series.
THE SELECTED ATLAS, consisting of 10 Maps, Imperial 410, 11
by 13 Inches, Stiff Cover, I 6
THE PORTABLE ATLAS, consisting of 16 Maps, folded Imperial
8vo, cloth lettered, % o
THE ADVANCED ATLAS, consisting of 32 Maps, Imperial 4to,
cloth lettered, ... ... ... ... ... ... ... 3 6
THE ACADEMIC ATLAS, consisting of 32 Maps, Imperial 4to,
with a Copious Index, cloth lettered, ... 50
THE STUDENT*S ATLAS, consisting of 32 Maps, and 6 Ancient
Maps, with a Copious Index, Imperial 8vo, cloth lettered, ... 6 o
THE COLLEGIATE ATLAS, consisting of 32 Modern, 16 Historical,
and 2 Ancient Maps, mounted on Guards, with a Copious Index,
Imperial 8vo, cloth lettered, ... 7 6
THE INTERNATIONAL ATLAS, consisting of 32 Modem, 16
Historical, and 14 Maps of Classical Geography, with Descriptive
Letterpress on Historical Geography, by W. F. Collier, LL.D.; and
on Classical Geography, by L. Schmitz, LL.D., with Copious Indices,
Imperial 8vo, Cloth mounted on Guards, ... ... ... ^ xo 6
PHYSICAL GEOGRAPHY— Demy Series.
THE PRIMARY ATLAS OF PHYSICAL GEOGRAPHYp 16
Maps, Demy 4to, 9 by 11 inches. Stiff Cover, I o
THE POCKET ATLAS OF PHYSICAL GEOGRAPHY, 16 Maps,
on Guards, Demy Svo, cloth, ... ... z o
liondoa, Bdinb'axB'bi, wi^'H.ekTtivoX.'SSCL^^^i'ka^^aiaagow.
'William Collins, Sons, & Co.'s Educational "Works.
COLLINS^ SERIES OF SCHOOL ATLASES—CoEtmuecL
HISTORICAL GEOGRAPHT.
THE POCKET ATLAS OF HISTORICAL GEOGRAPHY, i6
Maps, 6^ by ii inches, mounted on Guards, Imperial i6mo, cloth;
THE CROWN ATLAS OF HISTORKIAL GEOGRAPHY, 16
Maps, with Letterpress Description by Wmu F. Collier, LL.D.,
Imperial i6mo, cloth,
THE STUDENT*S ATLAS OF HISTORICAL GEOGRAPHY,
16 Maps, with Letterpress Description by Wm. F. Collier, LL.D.,
ovo, clOLn, ... ... ,,, ,,, .,, ... ••• •••
I
6
1 Roman Empire, Eastern and Western,
4th Century.
2 Europe, 6th Century, shewing Settle-
ments of the Barbarian Tribes.
3 Europe, 9th Century, shewing Empire
of Charlemagne.
4 Europe, 10th Century, at the Rise of
the Gennan Empire.
5 Europe, 12th Century, at the Time of
the Crusaders.
6 Europe, 16th Century, at the Eve of
the Reformation
7 Germany, 16th Century, Reformation
and Thirty YeaiV War.
8 Europe, 17th and 18th Centuries.
9 Europe at the Peace of 1815.
10 Europe in 1870.
11 India, illu.strating the Rise of the
British Empire.
12 World, on Mercator's Projection,
shewing Voyages of Discovery.
13 Britam under the Romans.
14 Britain under the Saxons.
15 Britain after Accession of William
the Conqueror.
16 France and Belgium, illustrating
British History.
CLASSICAL GEOGRAPHY.
THE POCKET ATLAS OF CLASSICAL GEOGRAPHY, 15
Maps, Imperial i6mo, 6^ by 11 inches, cloth lettered,
THE CROWN ATLAS OF CLASSICAL GEOGRAPHY. 15 Maps,
with Descriptive Letterpress, by Leonhard Schmitz, LL.D., Imperial
i6mo, cioth lettered, ... ... ... ...
THE STUDENT'S ATLAS OF CLASSICAL GEOGRAPHY, 15
Maps, Imperial 8vo, with Descriptive Letterpress, by Leonhard
Schmitz, LL.D., cloth lettered,
9 Armenia, Mesopotamia, &c
10 Asia Minor.
11 Palestine, (Temp. ChristL)
12 Gallia.
1 Orbis Veteribus Notus.
2 /Egyptus.
3 Regnum Alezandri Magni.
4 Macedonia, Thracia, &c.
5 Imperium Romanum.
*> Graecia.
7 Itaha, (Septentrionalis.)
8 Italia, (Meridionalis.)
13 Hispania.
14 Germania, &&
15 Britannia.
Historical and Classical Atlas.
THE STUDENT'S ATLAS OF HISTORICAL AND CLASSI-
CAL GEOGRAPHY, consisting of 30 Maps as above, with Intro-
ductions on Historical Geography by W. F. Collier, LL.D., and on
Classical Geography by Leonhard Schmitz, LL.D., with a Copijus
Index, Imperial 8vo, cloth.
Iiondon, Edinburgh, and HeTT\o\.'HXa.*^^tycV%^QS^»»«=s^
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COLLINS' SERIES OF SCHOOI. ITLISES— Contiimeft.
SGRIPTURK GEOGRAPHY.
THE ATLAS OF SCRIPTURE GtO<GRAPHY, i6 Maps, with s. d.
Questions on each Map, Stiff" Cover, ... ... ... ... i o
THE POCKhT ATLAS OF SCRIPTURE GEOGRAPHY, i6
Maps, 7i by 9 inches, mounted on; Guards, Imp. i6mo, cloth, ... i o
1 Ancient World, shewing probable Sclr
tlements of Descendants of Noali.
2 Counti-ies mentioned in the St.'iiptures,
3 Canaan in the time of the Patriarchs.
4 Journeyings of tiie Israelites.
5 Canaan as Divided among the Twelve
Tnbes.
6 The Donimions of David and Solomon.
7 Babylonii). A.ssyna, Media, and Susiana.;
9 Modem P?destine.
10 Physical Map of Palestine.
11 Journeys of the Apostle Paul.
12 Map shewing the prevailing Religions
of the World.
13 The Tabernacle in the Wilderness.
14 Plans of iSolomon's and Herod's Tem-
ples.
15 Plan of Ancient Jerusalem.
8 Palestine in the Time of Christ. I 16 Plan of Modern Jerusalem.
BLANK PR0JEX;TI0NS AND OUTLINES.
THE CROWN ATLAS OF BLANK PROJECTIONS, consisting
of 1 6 Maps, Demy 4to, on Stout Drawing Paper, Stiff" Wrapper, ... o 6
THE CROWN OUTLINE ATLAS, i6 Maps, Demy 4to, Stout
Drawing Paper, Stiff" Wrapper, o 6
THE IMPERIAL ATLAS OF BLANK PROJFXTIONS, consisting
of 1 6 Maps, Imperial 4to, on Stout Drawing Paper, Stiffs Wrapper, i 6
THE IMPERIAL OUTLINE ATLAS, i6 Maps, Imperial 410, Stout
Drawing Paper, Stiff" Cover, ... ... i 6
A Specimen Map of any of the foregoing Atlatafree on receipt oft*wo Penny Stamps.
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Printed in Colours^ and Mounted on Cloth and RolUn^ Garnished.
CHART OF THE WORLD, 5 ft. 2 in. by 4 ft. 6 in., 20 o
CENTRAL AND SOUTHERN EUROPE, 5 ft. 2 in. by 4 ft. 6 in., 20 o
EUROPE, ASIA, AFRICA, NORTH AMERICA, SOUTH
AMERICA, ENGLAND, SCOTLAND, IRELAND, PALES-
TINE, INDIA, each 3 ft. by 2 ft. 5 in., 6 6
UNITED STATES OF AMERICA, 3 ft. 11 inches by 2 ft. 4 in., 8 6
COUNTY WALL MAPS.
Printed in Colours, and Mounted on Cloth and Rollers^ Garnished.
MIDDLESEX, LANCASHIRE, YORKSHIRE, WARWICK,
DURHAM, CUMBERLAND, DERBYSHIRE, DORSET,
GLOUCESTER, HAMPSHIRE, SOMERSET, STAFFORD,
AND WILTS J each 54 in. by 48 in., 9 o
CHART OF METRIC SYSTEM.
CHART OF THE METRIC SYSTEM OF WEIGHTS AND
MEASURES. SUc 45 in. by 42 in., price, on Rollers, ... 9
TiOndon, EdinbixrgU, «ijCLeL'H.wrc\o\.'SLSSi^=K^^>8&,<i\«A€ow.
V.
r
♦*
'William Collins, Sons, & Co.'s Bduoational Works.
COLLINS' SERIES OF FIRST-GLASS SCHOOL ATLASES,
Carefully Constructed and Engraved from the best and latest Authorities^ and
Beautifully Printed in Colours^ on Superfine Cream JVove Paper.
MODERN GEOGRAPHY— Grown Series.
MY FIRST ATLAS, consisting of 12 Maps, 9 inches by 7 J inches,
folded 8vo, in Neat Wrapper,
1 The Hemispheres. | 7 Enj^Iand and Wales.
2 Europe.
3 Asia.
4 Africa.
6 North America.
6 ^outh Americii.
THE PRIMARY ATLAS, consisting of 16 Maps, 9 inches by 7^
inches, 4to, Stiff Wrapper, ...
s.
o
6
8 Scotland.
9 Ireland.
10 Central Europe.
11 Aastralia.
12 Palestiue.
9 Ireland.
10 Centi-al Europe.
IL India.
12 Canada.
13 United States.
14 Australia.
15 New Zealand.
16 I'alestine.
1 The Hemispheres.
2 Europe.
3 Asia.
4 Africa.
6 North America.
£:k)uth America
7 England and Wales.
8 Scotland.
THE POCKET ATLAS, consisting of 16 Maps, folded in 8vo, and
mounted on Guards, cloth lettered,
THE JUNIOR, OR YOUNG CHILD'S ATLAS, consisting of 1 6
Maps, 4to, with 16 pp. of Questions on the Maps, in Neat Wrapper,
THE SCHOOL BOARD ATLAS, consisting of 24 Maps, Crown
4to, cloth limp,
THE PROGRESSIVE ATLAS, consisting of 32 Maps, 9 inches by
7.J inches, 4to, cloth lettered, ...
1 Tlie Hemispheres.
2 The World. (Mcn-ator's rrojectiou )
3 Europe.
4 Asia.
5 Africa.
6 North America.
7 South America.
8 England and Wales.
9 Scotland.
10 Ireland.
11 France.
12 Holland and Belgium.
13 Switzerland.
14 Spain and Portugal.
15 Italy.
16 Sweden, Norway, and Denmarlc.
THE CROWN ATLAS, consisting of 32 Maps, on Guards, with
Index, 8 vo, cloth lettered, ...
THE NATIONAL ATLAS, consisting of 32 Maps, 4to, with a
Copious Index, cloth lettered,
17 Gorman Empire.
18 Austria,
lu Russia in Europa.
20 Turkey in Europe, and Greece.
21 India.
:!2 Persia, Afghanistan, and Beloochis-
:!3 Turkey in Asia [tan.
:U Chinese Empire and Japan.
25 Arabia, Egypt, and Nubia.
2t) Palestine.
27 Dominion of Canada.
28 United States.
29 West Indies.
30 Australia.
31 New South Wales, Victoria, and*
32 New Zealand. [South Australia.
J
6%.
JCondon, Edinburg\i, aTieL-awrcio\.^\XV ^^xY^.Q^».^ ^ L^^> ^ V
'Willi&m Collins, Sons, & Co's Educational "Works.
COLLINS' SERIES OF SOHOOL ATLASES-CONTIMUED.
MODERN GEOGRAPHY— Imperial Series.
THE SELECTED ATLAS, consisting of 1 6 Maps, Imperial 4to, 1 1
by 13 inches, Stiff Cover,
1 The Hemispheres.
2 Europe.
3 Asia.
4 Africa.
5 Nortli America. •
6 South America.
7 England and Wales.
8 Scotland.
s.
••• ••• ••• ••• A
9 Ireland.
10 Southern and Central Europe.
11 India.
12 Canada.
13 United States.
14 Australia-
Id New Zealand.
16 Palestine.
6
THE PORTABLE ATLAS, consisting of 16 Maps, folded Imperial
8vo, cloth lettered,
THE ADVANCED ATLAS, consisting of/32 Maps, Imperial 4to.
cloth lettered,
1 Eastern and Western Hemispheres.
2 The World, (Mercator's Projection.)
3 Europe.
4 Asia.
5 Africa.
North America.
7 South Ameiica.
8 England and Wales.
9 Scotland.
10 Ireland.
11 France.
12 Holland and Belgium.
13 Switzerland,
14 Spain and PoiiugaL
1.3 Italy. [the Baltic.
10 Sweden and Norwaj', Denmark and
3 6
17 German Empire.
18 Austria.
19 Russia.
20 Turkey in Europe, and Greece.
21 India.
22 Persia, Afghanistan, and Beloochis-
1^3 Turkey in Asia. [tan.
24 Cliinese Empire, and Japan.
lij Arabia. Egj'pt, Nubia, and Abys-
20 Palestine. [sinia.
27 Dominion of Canada.
28 United States.
li) West Indies and Central America.
30 Australia.
31 Victfjria, New Soutli Wales, and
32 New Zealand. [South Australia.
THE ACADEMIC ATLAS, consisting of 32 Maps, Imperial 4to,
with a Copious Index, cloth lettered,
THE STUDENTS ATLAS, consisting of 32 Maps, and 6 Ancient
Maps, with a Copious Index, Imperial 8vo, cloth lettered^
":i Ancient Greece.
.'! I Ancient Roman Empire.
.'{•> Britain under the Romans.
30 Britain under the Saxons.
37 Historical Map of the British Is-
lands from A.D. 1066.
38 France and Belgium, illustrating
British History.
THE COLLEGIATE ATLAS, consisting of 32 Modern, 16 Historical,
and 2 Ancient Maps, mounted on Guards, with a Copiojs Index,
Imperial 8vo, cloth lettered,
'iHE INTERNATIONAL ATLAS, consisting of 32 Modern, 16
Historical, and 14 Maps of Classical Geography, with Descriptive
Letterpress on Historical Geognnhir hv W g i'^iiimmUmA^t^nd
7 6
on Classical Geography by L. S^
Imperial 8vo, cloth mounted ol
London, Edinhursh,
•**•
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